For the graph of $f\left(x\right)=2x^2-8x+1$

Find:

ⓐ the axis of symmetry

ⓑ the vertex.

For the graph of $f\left(x\right)=2x^2-8x+1$

We want to find:

ⓐ the axis of symmetry

ⓑ the vertex.

We know that Axis of Symmetry and Vertex of a Parabola:

The graph of the function $f\left(x\right)=a{x}^2+bx+c$is a parabola where:

• the axis of symmetry is the vertical line$x=-\frac{b}{2a}$
• the vertex is a point on the axis of symmetry, so its x-coordinate is$-\frac{b}{2a}$
• the y-coordinate of the vertex is found by substituting $x=-\frac{b}{2a}$into the quadratic equation.

Solution of the axis of symmetry

$$\begin{gathered} f\left(x\right)=a{x}^2+bx+c \\ f\left(x\right)=2x^2-8x+1 \end{gathered}$$

The axis of symmetry is the vertical line

$x=-\frac{b}{2a}$

Substitute the values of a, b into the

equation.

localid="1646149690675" $x=-\frac{-8}{2\left(2\right)}$

Simplify

$x=2$

The axis of symmetry is the line x = 2 .

$$\begin{gathered} f\left(x\right)=a{x}^2+bx+c \\ f\left(x\right)=2x^2-8x+1 \end{gathered}$$

The vertex is a point on the line of

symmetry, so its x-coordinate will be

x = 2 .

Find f (2) .

$$\begin{gathered} f\left(x\right)=2x^2-8x+1 \\ f\left(2\right)=2{\left(2\right)}^2-8\left(2\right)+1 \\ f\left(2\right)=8-16+1 \\ f\left(2\right)=-7 \end{gathered}$$

The vertex is (2, −7) .