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Q61.

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Found in: Page 293

### Algebra 2

Book edition Middle English Edition
Author(s) Carter
Pages 804 pages
ISBN 9780079039903

# Find the coordinates of the maximum or minimum value of each quadratic equation to the nearest hundredth.$f\left(x\right)=-6{x}^{2}+9x$

The coordinates of the minimum value of the quadratic equation $f\left(x\right)=-6{x}^{2}+9x$ to the nearest hundredth is $\left(0.75,3.375\right)$.

See the step by step solution

## Step 1. Given Information.

Given to determine the coordinates of the maximum or minimum value of the quadratic equation $f\left(x\right)=-6{x}^{2}+9x$ 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, $a=-6,b=9$

Plugging the values in the equation:

$\begin{array}{l}x=-\frac{b}{2a}\\ x=-\frac{\left(9\right)}{2\left(-6\right)}\\ x=\frac{9}{12}\\ x=0.75\end{array}$

Hence the x-coordinate of the vertex is $x=0.75.$

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

Plugging $x=0.75$ in the equation:

$\begin{array}{l}f\left(x\right)=-6{x}^{2}+9x\\ f\left(0.75\right)=-6{\left(0.75\right)}^{2}+9\left(0.75\right)\\ f\left(0.75\right)=-3.375+6.75\\ f\left(0.75\right)=3.375\end{array}$

Hence the coordinates of the vertex is $\left(0.75,3.375\right)$.