Q59. - Algebra 2

Question

Find the coordinates of the maximum or minimum value of each quadratic equation to the nearest hundredth.

 

$f\left(x\right)=-5x^2+8x$

Step-by-Step Solution

Answer

The coordinates of the minimum value of the quadratic equation $f\left(x\right)=-5x^2+8x$ to the nearest hundredth is $\left(0.8,3.2\right)$.

Step 1. Given Information.

Given to determine the coordinates of the maximum or minimum value of the quadratic equation $f\left(x\right)=-5x^2+8x$ to the nearest hundredth.

Step 2. Explanation .

The maximum or minimum value of a quadratic function lies at the vertex of the graph.

For an equation of the form $f\left(x\right)=a{x}^2+bx+c$:

if a>0, it is an upwards opening parabola and so has a minimum value

if a<0, it is a downwards opening parabola and so has a maximum value.

So, the given quadratic equation has a maximum value.

 

For an equation of the form $f\left(x\right)=a{x}^2+bx+c$, the x-coordinate of the vertex is given by $x=-\frac{b}{2a}$

Here for the given equation, a=-5,b=8

Plugging the values in the equation:

 $\begin{array}{l}x=-\frac{b}{2a}\\ x=-\frac{\left(8\right)}{2\left(-5\right)}\\ x=\frac{8}{10}\\ x=0.8\end{array}$

Hence the x-coordinate of the vertex is x=0.8.

 

So, the y-coordinate of the vertex can be obtained by plugging the value in the equation.

Plugging x=0.8 in the equation:

 $\begin{array}{l}f\left(x\right)=-5x^2+8x\\ f\left(0.8\right)=-5{\left(0.8\right)}^2+8\left(0.8\right)\\ f\left(0.8\right)=-5\left(0.64\right)+8\left(0.8\right)\\ f\left(0.8\right)=-3.2+6.4\\ f\left(0.8\right)=3.2\end{array}$

Hence the coordinates of the vertex is $\left(0.8,3.2\right)$.

Step 3. Conclusion .

The coordinates of the maximum value of the quadratic equation $f\left(x\right)=-5x^2+8x$ to the nearest hundredth is $\left(0.8,3.2\right)$.