Let us recall the graph of a quadratic function. Do you remember the name of the U-shaped curve it produces? That's right! It's called a parabola.

Parabola example, StudySmarter Originals

Now, notice how the graph curves down and hits a point where it then curves up. We see that the parabola encounters a change in direction upon meeting this exact location. This is called the turning point.

In this lesson, we shall look at the idea of a turning point and introduce several methods we can use to determine the turning point of a given function. Throughout this topic, we will only focus on finding the turning point of a quadratic function of the form,

$y=a{x}^{2}+bx+c$.

## The Meaning of a Turning Point

As introduced at the beginning of this article, the turning point of a graph is a point at which the curve deviates from its initial path. That is to say, if a car was driving up a hill and reached its peak, it will then have to drive down in order to get to the other side of the hill. Below is a formal definition of a turning point.

The **turning point **of a graph is a point at which the curve changes direction.

In some textbooks, a turning point may also be referred to as a critical point.

### A Turning Point on a Graph

The turning point of a curve is represented by a pair of x and y-coordinates on the Cartesian plane and is denoted by the point, (x, y). Below is an example of what a turning point may look like when it is plotted on a graph.

The turning point on the Cartesian plane, StudySmarter Originals

### Types of Turning Points

There are two types of turning points we need to consider when graphing quadratic functions:

**Minimum Point,**and**Maximum****Point**.

Each type has its own orientation of curvature and can be determined by looking at the value of the leading coefficient of the quadratic function, i.e. the coefficient of x^{2}. The table below describes the key features of both these cases along with their general plot.

Minimum Point | Maximum Point |

The coefficient of x | The coefficient of x |

The minimum point, StudySmarter Originals | The maximum point, StudySmarter Originals |

The graph curves down reaching its minimum and then curves up | The graph curves up reaching its maximum and then curves down |

The turning point is the lowest point of the curve | The turning point is the highest point of the curve |

The y-value of the turning point is the smallest possible value of the function | The y-value of the turning point is the largest possible value of the function |

## Turning Point Formula

The turning point of a quadratic equation coincides with the vertex of its corresponding parabola. To determine the coordinates of its vertex, we shall make use of the **Vertex Formula**, or in this case, the **Turning Point Formula**. Given the standard form of a quadratic equation,

$y=a{x}^{2}+bx+c$

the vertex form of this quadratic equation is

$y=a{(x-h)}^{2}+k$,

where (h, k) is the vertex or turning point of the parabola. There are two ways in which we can determine the values of h and k.

### Method 1

Evaluate $h=-\frac{b}{2a}$ and $k=-\frac{D}{4a}$ where D is the discriminant and $D={b}^{2}-4ac$.

### Method 2

Calculate $h=-\frac{b}{2a}$ and use this result to find k by substituting this value of h into y.

Find the turning point of the quadratic equation below.

$y=3{x}^{2}-8x+1$

**Solution**

Let us use the first method of the turning point formula to find the values of h and k.

Here, a = 3, b = –8 and c = 1. The value of h is given by,

$h=-\frac{(-8)}{2\left(3\right)}\Rightarrow h=\frac{4}{3}$

The discriminant is given by,

$D={(-8)}^{2}-4\left(3\right)\left(1\right)\Rightarrow D=52$

Thus, k is given by,

$k=-\frac{52}{4\left(3\right)}\Rightarrow k=-\frac{13}{3}$

Hence, the turning point is $\left(\frac{4}{3},-\frac{13}{3}\right)$.

**Verify**

Let us check whether our solution is correct by using the second method of the turning point formula. As before, we know that

$h=-\frac{(-8)}{2\left(3\right)}\Rightarrow h=\frac{4}{3}$

Substituting this value of h into y, we obtain

$k=y\left(\frac{4}{3}\right)\Rightarrow k=3{\left(\frac{4}{3}\right)}^{2}-8\left(\frac{4}{3}\right)+1\Rightarrow k=-\frac{13}{3}$

Thus, our turning point is $\left(\frac{4}{3},-\frac{13}{3}\right)$ , as required.

Note that this formula can only be applied to quadratic equations as in the form above. It is not applicable to finding turning points of higher degree polynomials and other complex functions. However, this is where the next section will be of relevance! Here, we shall look into other ways in which we can locate turning points of any type of equation.

## The Math of Finding the Turning Point

For a given quadratic function, we can locate the turning point of the parabola using four methods, namely:

Line of symmetry

Factorisation

In this segment, we shall provide a detailed step-by-step process of each technique stated above along with several worked examples.

### Line of Symmetry Method

We shall begin by observing a distinct trait of a parabola that allows us to determine its turning point. This property is known as its line of symmetry. Recall that the line of symmetry is a line segment that divides an object precisely in half.

The graph of any quadratic function has a **vertical line of symmetry** that passes through its turning point. That is to say, the turning point is situated exactly in the middle of the parabola. To identify the turning point of a given curve, we can use the following steps.

** Step 1:** Determine whether the graph is a minimum or maximum by observing the coefficient of x^{2};

** Step 2:** Identify the equation for the line of symmetry using the formula,

This is the x-coordinate of the turning point;

**Step 3: **Evaluate the y-coordinate by substituting the x-coordinate from Step 3 back into the given function.

Below are two worked examples applying this method.

Find the turning point of the quadratic equation below using the line of symmetry method.

$f\left(x\right)=2{x}^{2}-3x-1$

**Step 1:** Looking at the coefficient of x^{2}, we have a = 2 > 0. Since a is positive, the turning point of this curve must be a minimum.

**Step 2:** The x-coordinate of the turning point is given by the equation for the line of symmetry. Here, a = 2 and b = –3. Then,

**Step 3: **Substituting this value of x into our original equation, we obtain our y-coordinate as

$f\left(\frac{3}{4}\right)=2{\left(\frac{3}{4}\right)}^{2}-3\left(\frac{3}{4}\right)-1\Rightarrow f\left(\frac{3}{4}\right)=-\frac{17}{8}$

Thus, the turning point is a minimum point given by $\left(\frac{3}{4},-\frac{17}{8}\right)$. As this is a minimum point, the graph curves downwards. This is shown below.

Example 1, StudySmarter Originals

Using the line of symmetry method, identify the turning point of the following quadratic equation.

$f\left(x\right)=-{x}^{2}+4x-2$

**Step 1:** Looking at the coefficient of x^{2}, we have a = -1 < 0. Since a is negative, the turning point of this curve must be a maximum.

**Step 2:** The x-coordinate of the turning point is given by the equation for the line of symmetry. Here, a = –1 and b = 4. Then,

**Step 3: **Substituting this value of x into our original equation, we obtain our y-coordinate as

$f\left(2\right)=-{\left(2\right)}^{2}+4\left(2\right)-2\Rightarrow f\left(2\right)=2$

Thus, the turning point is a maximum point given by $\left(2,2\right)$. As this is a maximum point, the graph curves upwards. This is shown below.

Example 2, StudySmarter Originals

### Factorisation Method

Given that a vertical line of symmetry cuts the parabola into two equal halves and crosses its turning point, we can say that the position of its turning point is exactly at the midpoint between the (two) x-intercepts of a quadratic function. With that in mind, we can find the turning point of a quadratic equation with an approach involving **factorisation**. There are five steps to this method.

** Step 1: **Determine whether the graph is a minimum or maximum by observing the coefficient of x^{2};

** Step 2:** Factorise the given quadratic function and set it to zero;

** Step 3:** Identify the roots of the function;

** Step 4:** Locate the x-coordinate of the turning point by taking the midpoint of the pair of roots;

**Step 5: **Evaluate the y-coordinate by substituting the x-coordinate from Step 4 back into the given function.

Let us now look at some worked examples that demonstrate this technique.

Find the turning point of the quadratic equation below using the factorisation method.

$f\left(x\right)=-2{x}^{2}-12x-10$

**Step 1: **Looking at the coefficient of x^{2}, we have a = –2 < 0. Since a is negative, the turning point of this curve must be a maximum.

**Step 2:** Factorising the quadratic function and setting it to zero, we obtain

$f\left(x\right)=0\Rightarrow -2{x}^{2}-12x-10=0\Rightarrow (-x-5)(2x+2)=0$

**Step 3:** The roots of the function are given by

$-x-5=0\Rightarrow x=-5$

and

$2x+2=0\Rightarrow 2x=-2\Rightarrow x=-1$

Thus the roots are x = –1 and x = –5.

**Step 4:** To locate the midpoint, M, of these two x-coordinates, we simply find their average

$M=\frac{-1+(-5)}{2}\Rightarrow M=-\frac{6}{2}\Rightarrow M=-3$

The x-coordinate of our turning point is M or x = –3.

**Step 5: **Substituting this value of x into our original equation, we obtain our y-coordinate as

$f(-3)=-2{(-3)}^{2}-12(-3)-10\Rightarrow f(-3)=8$

Thus, the turning point is a maximum point given by $\left(-3,8\right)$. As this is a maximum point, the graph curves upwards. This is shown below.

Example 3, StudySmarter Originals

Using the factorisation method, identify the turning point of the following quadratic equation.

$f\left(x\right)={x}^{2}-3x-18$

**Step 1: **Looking at the coefficient of x^{2}, we have a = 1 > 0. Since a is positive the turning point of this curve must be a minimum.

**Step 2:** Factorising the quadratic function and setting it to zero, we obtain

$f\left(x\right)=0\Rightarrow {x}^{2}-3x-18=0\Rightarrow (x+3)(x-6)=0$

**Step 3:** The roots of the function are given by

$x+3=0\Rightarrow x=-3$

and

$x-6=0\Rightarrow x=6$

Thus the roots are x = –3 and x = 6.

**Step 4:** To locate the midpoint, M, of these two x-coordinates, we simply find their average

$M=\frac{-3+6}{2}\Rightarrow M=\frac{3}{2}$

**Step 5: **Substituting this value of x into our original equation, we obtain our y-coordinate as

$f\left(\frac{3}{2}\right)={\left(\frac{3}{2}\right)}^{2}-3\left(\frac{3}{2}\right)-18\Rightarrow f\left(\frac{3}{2}\right)=-\frac{81}{4}$

Thus, the turning point is a minimum point given by $\left(\frac{3}{2},-\frac{81}{4}\right)$. As this is a minimum point, the graph curves downwards. This is shown below.

Example 4, StudySmarter Originals

### Completing the Square Method

Next, we shall now move on to finding the turning point of a quadratic equation using a method called **completing the square**. When we complete the square of a given quadratic function, it will take the form,

$y=a{(x-h)}^{2}+k$

To recall the process of this method, refer to the topic: Completing the square. There are 3 steps to consider in this case.

**Step 1: **Determine whether the graph is a minimum or maximum by observing the coefficient of x^{2};

**Step 2: **Complete the square of the given quadratic function so that it looks like $y=a{(x-h)}^{2}+k$;

**Step 3:** Identify the x and y-coordinates of the turning point. The value of h and k are the x and y-coordinates respectively.

Now that we have established the steps above, let us observe two worked examples.

Find the turning point of the quadratic equation below using the completing the square method.

$f\left(x\right)=2{x}^{2}+9x$

**Step 1: **Looking at the coefficient of x^{2}, we have a = 2 > 0. Since a is positive the turning point of this curve must be a minimum.

**Step 2: **Completing the square of the quadratic function, we obtain

$f\left(x\right)=2{x}^{2}+9x\Rightarrow f\left(x\right)=2\left({x}^{2}+\frac{9}{2}x\right)\Rightarrow f\left(x\right)=2{\left(x+\frac{9}{4}\right)}^{2}-2{\left(\frac{9}{4}\right)}^{2}\Rightarrow f\left(x\right)=2{\left(x+\frac{9}{4}\right)}^{2}-\frac{81}{8}$

Note that we need to factor out the coefficient of x^{2}_{, }halve the coefficient of x to complete the square and balance the equation. Now our function looks like the form $y=a{(x-h)}^{2}+k$.

**Step 3:** From here, $h=-\frac{9}{4}$ and $k=-\frac{81}{8}$. The values of h and k represent the x and y-coordinates of the turning point, respectively.

Thus, the turning point is a minimum point given by $\left(-\frac{9}{4},-\frac{81}{8}\right)$. As this is a minimum point, the graph curves downwards. This is shown below.

Example 5, StudySmarter Originals

Using the completing the square method, identify the turning point of the following quadratic equation.

$f\left(x\right)=-{x}^{2}+7x+4$

**Step 1: **Looking at the coefficient of x^{2}, we have a = –1 < 0. Since a is negative, the turning point of this curve must be a maximum.

**Step 2: **Completing the square of the quadratic function

$f\left(x\right)=-{x}^{2}+7x+4\Rightarrow f\left(x\right)=-\left({x}^{2}-7x\right)+4\Rightarrow f\left(x\right)=-{\left(x-\frac{7}{2}\right)}^{2}+4+{\left(\frac{7}{2}\right)}^{2}\Rightarrow f\left(x\right)=-{\left(x-\frac{7}{2}\right)}^{2}+\frac{65}{4}$

Note that we need to factor out the coefficient of x^{2}_{, }halve the coefficient of x to complete the square and balance the equation. Now our function looks like the form $y=a{(x-h)}^{2}+k$.

**Step 3: **From here, $h=\frac{7}{2}$ and $k=\frac{65}{4}$. The values of h and k represent the x and y-coordinates of the turning point, respectively.

Thus, the turning point is a maximum point given by $\left(\frac{7}{2},\frac{65}{4}\right)$. As this is a maximum point, the graph curves upwards. This is shown below.

Example 6, StudySmarter Originals

### Differentiation Method

Let us now look at our final method of locating the turning point of a quadratic function. This time, we shall be using a technique widely used in calculus called **differentiation**. Recall that differentiation is a method used to determine the gradient of a curve at a specific point. A more detailed overview of this topic can be found here: Differentiation.

For a quadratic function of the form$y=a{x}^{2}+bx+c$, the gradient of the curve at its turning point equals zero. In other words, $\frac{dy}{dx}=0$.

As mentioned before, a turning point can either be a minimum or maximum. Let us look at each case in turn.

Minimum Point | Maximum Point |

The gradient of minimum point, StudySmarter Originals | The gradient of a maximum point, StudySmarter Originals |

At a minimum point, the gradient changes from negative to zero to positive. | At a maximum point, the gradient changes from positive to zero to negative. |

To locate the turning point using differentiation, simply calculate the first derivative of the given function and set it to zero. This will give you the x-coordinate of the turning point. The y-coordinate is found by substituting this found x-value into the initial quadratic function.

$\frac{dy}{dx}=0$ or $f\text{'}\left(x\right)=0$

The nature of the turning point is found by calculating the second derivative of the given function. Then,

if $\frac{{d}^{2}y}{{d}^{2}x}>0$ or $f\text{'}\text{'}\left(x\right)>0$ then the turning point of our function is a minimum;

if $\frac{{d}^{2}y}{{d}^{2}x}<0$ or $f\text{'}\text{'}\left(x\right)<0$ then the turning point of our function is a maximum;

if $\frac{{d}^{2}y}{{d}^{2}x}=0$ or $f\text{'}\text{'}\left(x\right)=0$ then the turning point of our function is an inflection point.

Note that the third clause will be further discussed on the topic of cubic functions and higher-order polynomials.

As we now have these facts covered, let us now observe two examples.

Use differentiation to locate the turning points of the following quadratic equation.

$f\left(x\right)=-4{x}^{2}-x$

**Solution **

First, let us locate the turning points of the curve. Calculating the first derivative of the function and setting it to zero, we obtain

$f\text{'}\left(x\right)=0\Rightarrow -8x-1=0\Rightarrow -8x=1\Rightarrow x=-\frac{1}{8}$

The y-coordinate is given by

$f\left(-\frac{1}{8}\right)=-4{\left(-\frac{1}{8}\right)}^{2}-\left(-\frac{1}{8}\right)f\left(-\frac{1}{8}\right)=\frac{1}{16}$

Thus, the turning point is $\left(-\frac{1}{8},\frac{1}{16}\right)$

The nature of this turning point is given by the second derivative.

$f\text{'}\text{'}\left(x\right)=-8<0$

Since the value of the second derivative is negative, then our turning point is a maximum. Thus, the graph curves upwards. This is shown below.

Example 7, StudySmarter Originals

Given the quadratic equation below, determine its turning point using differentiation.

$f\left(x\right)=2{x}^{2}+7x+9$

**Solution **

First, let us locate the turning points of the curve. Calculating the first derivative of the function and setting it to zero, we obtain

$f\text{'}\left(x\right)=0\Rightarrow 4x+7=0\Rightarrow 4x=-7\Rightarrow x=-\frac{7}{4}$

The y-coordinate is given by

$f\left(-\frac{7}{4}\right)=2{\left(-\frac{7}{4}\right)}^{2}+7\left(-\frac{7}{4}\right)+9\Rightarrow f\left(-\frac{7}{4}\right)=\frac{23}{8}$

Thus, the turning point is $\left(-\frac{7}{4},\frac{23}{8}\right)$ .

The nature of this turning point is given by the second derivative.

$f\text{'}\text{'}\left(x\right)=4>0$

Since the value of the second derivative is positive, then our turning point is a minimum. Thus, the graph curves downwards. This is shown below.

Example 8, StudySmarter Originals

## Turning Point Examples

To further enhance our skills in finding the turning point of a parabola, we shall end our discussion with two real-world examples involving the concept of a turning point.

The trajectory of a tennis ball thrown across a court is modelled by

$h\left(x\right)=-0.3{x}^{2}-2.7x+1.8$,

where h(x) represents the height of the ball from the ground in meters. What is the maximum height the ball can reach based on this projectile model?

Example 9, StudySmarter Originals

**Solution **

We shall use the differentiation method to solve this problem. As mentioned in the question, we are looking for a maximum turning point, which is the peak of the parabola. This is indeed a maximum point as the coefficient of x^{2} is negative since a = –0.3. As seen above, our parabola curves upwards, as required.

To locate the turning point of this curve, we shall first calculate the first derivative of the given model and set it to zero.

$h\text{'}\left(x\right)=0\Rightarrow -0.6x-2.7=0$

Solving this, we obtain

$-0.6x=2.7\Rightarrow x=-9.2$

This x-coordinate dictates the position at which the ball is at its highest point. Now plugging this value of x into our original equation, we have

$h(-9.2)=-0.3{(-9.2)}^{2}-2.7(-9.2)+1.8\Rightarrow h(-9.2)=7.875$

Thus, the highest point that the ball can reach from the ground is 7.875 meters.

The motion of a pendulum follows a parabolic curve modelled by

$p\left(x\right)=0.22{x}^{2}+x-1.6$,

where p(x) is the length of the pendulum from its point of suspension in centimetres. Determine the length of the pendulum from the bar when the pendulum string is at right angles from the point of suspension.

Example 10, StudySmarter Originals

**Solution **

We shall use the line of symmetry method to tackle this problem. First of all, note that the coefficient of x^{2} is a = 0.22 > 0. Since this is positive, our turning point is a minimum. As you can see above, the parabola curved downwards, as required. Here, b = 1. Applying the formula for the equation of the line of symmetry, we obtain

$x=-\frac{b}{2a}\Rightarrow x=-\frac{1}{2(0.22)}\Rightarrow x=-\frac{25}{11}$

Substituting this value of x into our original equation yields

$p\left(\frac{25}{11}\right)=0.22{\left(\frac{25}{11}\right)}^{2}+\left(\frac{25}{11}\right)-1.6\Rightarrow p\left(\frac{25}{11}\right)=-\frac{301}{110}\Rightarrow p\left(\frac{25}{11}\right)\approx -2.74(correctto2decimalplaces)$

Since we are dealing with lengths, we can take this value as positive as there is no such thing as a negative distance. Thus, the length of the pendulum from the bar when the pendulum string is at right angles from the point of suspension is 2.74 centimetres.

**Try it yourself: **Attempt using the other methods for finding the turning points for these two questions above. Do they give you the same answer?

## Turning Points - Key takeaways

- The turning point of a graph is a point at which the curve changes direction.
- The turning point of a graph is denoted by the coordinates (x, y).
- For a quadratic function of the form y = ax
^{2}+ bx + c,- If a < 0 the turning point is a minimum and is the lowest point of the curve.
- If a > 0 the turning point is a maximum and is the highest point of the curve.

- Methods to find the turning point of a graph
- Line of symmetry
- Factorisation
- Completing the square
- Differentiation

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##### Frequently Asked Questions about Turning Points

How do you find the turning point?

You can find the turning points via:

- Line of symmetry method
- Factorisation method
- Completing the square method
- Differentiation method

What is a turning point on a graph?

The turning point of a graph is a point at which the curve changes direction.

How do you write a turning point in math?

The turning point of a graph is denoted by the coordinates (x, y).

How to find the turning point of a quadratic equation?

You can find the turning points of a quadratic formula using:

- Line of symmetry method
- Factorisation method
- Completing the square method
- Differentiation method

What is a turning point in math?

The turning point of a graph is a point at which the curve changes direction.

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