Question

In the following exercises, graph by using intercepts, the vertex, and the axis of symmetry.

$y=-4x^2+16x-11$

Short answer

The given information is:

$y=-4x^2+16x-11$

Step-by-step solution

Step 1. Given information.

The given information is:

$y=-4x^2+16x-11$

Step 2. Find the axis of symmetry and the vertex.

$$\begin{gathered} y=a{x}^2+bx+c \\ y=-4x^2+16x-11 \end{gathered}$$

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.

$$\begin{gathered} x=-\frac{16}{2\left(-4\right)} \\ =2 \end{gathered}$$

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,

$$\begin{gathered} y=-4{\left(2\right)}^2+16\left(2\right)-11 \\ =-4\left(4\right)+32-11 \\ =5 \end{gathered}$$

Therefore, the vertex is$\left(2,5\right)$.

Step 3. Find the intercepts.

Now substitute xequal to 0 in the equation to find the intercept y,

$$\begin{gathered} y=-4{\left(0\right)}^2+16\left(0\right)-11 \\ =-11 \end{gathered}$$

Therefore, the point $\left(0,-11\right)$ is the y-intercept.

We can see that the resultant point is left of the line of symmetry by 2 units.

Therefore the point right to the line of symmetry by 2 units is $\left(4,-11\right)$.

Now substitute yequal to 0 in the equation to find the intercept x,

role="math" localid="1653846749886" $$\begin{gathered} -4x^2+16x-11=0 \\ x=\frac{-16\pm \sqrt{{\left(16\right)}^2-4\left(-4\right)\left(-11\right)}}{2\left(-4\right)} \\ x=\frac{16\pm \sqrt{80}}{-8} \\ x=\frac{-4\pm \sqrt{15}}{-2} \end{gathered}$$

Therefore, the points of the x-intercept are$\left(\frac{-4-\sqrt{15}}{-2},0\right)\mathrm{and}\left(\frac{-4+\sqrt{15}}{-2},0\right)$.

Step 4. Plot the graph.