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For the graph of$f\left(x\right)=2x^2鈭�4x鈭�3$

We want to find:

鈸� the axis of symmetry

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

$f\left(x\right)=2x^2鈭�4x鈭�3$

The axis of symmetry is the vertical line

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

Substitute the values of a, b into the

equation.

localid="1646150751313" $x=-\frac{-4}{2\left(2\right)}$

Simplify:

$x=1$

The axis of symmetry is the line x = 1 .

$f\left(x\right)=2x^2鈭�4x鈭�3$

The vertex is a point on the line of symmetry, so its x-coordinate will be x = 1.

Find f(1):

$$\begin{gathered} f\left(x\right)=2x^2鈭�4x鈭�3 \\ f\left(1\right)=2{\left(1\right)}^2鈭�4\left(1\right)鈭�3 \\ f\left(1\right)=2鈭�4鈭�3 \\ f\left(1\right)=-5 \end{gathered}$$

The vertex is (1, 鈭�5) .