Question

What is the x intercept, y intercept, and vertex of equation:e(x)=-2x^2+8x-1

Answer 1

Given the quadratic equation as shown below

$f(x)=-2x^2+8x-1$$\therefore y=-2x^2+8x-1$

To find the y intercept, set x=0 and solve for y

$\begin{gathered} x=0 \\ y=-2(0^2)+8(0)-1 \end{gathered}$$\begin{gathered} y=0+0-1=-1 \\ \text{Thus, y-intercept is}\Rightarrow(0,-1) \end{gathered}$

To find the x-intercepts, set y=0 and solve for x

$\begin{gathered} y=0 \\ 0=-2x^2+8x-1 \\ -2x^2+8x-1=0 \\ \text{Using quadratic formula} \\ a=-2,b=8,c=-1 \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ x=\frac{-8\pm\sqrt[]{8^2-4\times(-2)\times(-1)}}{2\times(-2)} \end{gathered}$$\begin{gathered} x=\frac{-8\pm\sqrt[]{64-8}}{-4}=\frac{-8\pm\sqrt[]{56}}{-4} \\ x=\frac{-8\pm2\sqrt[]{14}}{-4} \end{gathered}$$x=\frac{-8+2\sqrt[]{14}}{-4}\text{ OR }\frac{-8-2\sqrt[]{14}}{-4}$$\begin{gathered} x=\frac{-8}{-4}+\frac{2\sqrt[]{14}}{-4}\text{ OR }\frac{-8}{-4}-\frac{2\sqrt[]{14}}{-4} \\ x=2-\frac{1}{2}\sqrt[]{14}\text{ OR }2+\frac{1}{2}\sqrt[]{14} \\ \text{The x-intercepts are} \\ (2-\frac{1}{2}\sqrt[]{14},0)\text{ and (}2+\frac{1}{2}\sqrt[]{14},0) \end{gathered}$

The vertex of the equation is given by

$\begin{gathered} y=-2x^2+8x-1 \\ y=-2(x^2-4x)-1 \\ y=-2\lbrack(x-2)^2-4\rbrack-1\text{ ( by completing the square of the quadratic binomial)} \\ y=-2(x-2)^2+8-1 \\ y=-2(x-2)^2+7 \end{gathered}$

Given the vertex form equation as shown below

$\begin{gathered} y=a(x-h)^2+k \\ \text{Vertex}\Rightarrow(h,k) \end{gathered}$

From the equation

$\begin{gathered} y=-2(x-2)^2+7 \\ h=2 \\ k=7 \end{gathered}$

Thus, the vertex of the function is

$(2,7)$

Hence, the summary of the answer is given below

$\begin{gathered} y-\text{intercept}\Rightarrow(0,-1) \\ x-\text{intercepts}\Rightarrow(2-\frac{1}{2}\sqrt[]{14},0)\text{ and (}2+\frac{1}{2}\sqrt[]{14},0)\text{ OR }(0.129,0)\text{ and }(3.871,0) \\ \text{Vertex}\Rightarrow(2,7) \end{gathered}$