Turning Points

Find the turning point of the quadratic equation below.

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

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.

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

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

Step 1: Looking at the coefficient of x², 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 x², 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

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

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

Step 1: Looking at the coefficient of x², 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

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

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 x², 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

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

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

Step 1: Looking at the coefficient of x², 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

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

Note that we need to factor out the coefficient of x²_, 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 x², 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 x²_, 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

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

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

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.

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

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

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

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

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 x² 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

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

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 x² 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.