Question

In the following exercises, find the minimum or maximum value.

$y=-3x^2+12x-10$

Short answer

The maximum value is 2.

Step-by-step solution

Step 1. Given information.

The given information is:

$y=-3x^2+12x-10$

Step 2. Find the maximum value of the function.

$$\begin{gathered} y=a{x}^2+bx+c \\ y=-3x^2+12x-10 \end{gathered}$$

The value of coefficient ais negative therefore we can say that the parabola opens down.

The axis of symmetry is the line $x=-\frac{b}{2a}$.

Substitute the values of a, and binto the equation.

$$\begin{gathered} x=-\frac{12}{2\left(-3\right)} \\ =2 \end{gathered}$$

Therefore, the axis of symmetry is $x=2$.

The vertex is on the line of symmetry, so its x-coordinate will be $x=2$.

Now substitute the value of xinto the equation,

$$\begin{gathered} y=-3{\left(2\right)}^2+12\left(2\right)-10 \\ =-12+24-10 \\ =2 \end{gathered}$$

Therefore, the vertex is (2,2).

The maximum value of the given function is 2 and it occurs at the value of xas 2.