# How do you graph the quadratic function and identify the vertex and axis of symmetry for y=-x^2+4x-2?

Mar 21, 2017

Vertex$\to \left(x , y\right) = \left(2 , 2\right)$
Axis of symmetry $\to x = 2$

#### Explanation:

As the ${x}^{2}$ part is negative the graph is of general shape $\cap$

The axis of symmetry coincides with ${x}_{\text{vertex}}$

$\textcolor{b l u e}{\text{Determine the vertex and axis of symmetry}}$

Write as $y = - 1 \left(x \textcolor{red}{- 4} x\right) - 2$

"Axis of symmetry "=x_("vertex")=(-1/2)xx(color(red)(-4)) = +2

By substitution:

${y}_{\text{vertex}} = - {\left(2\right)}^{2} + 4 \left(2\right) - 2 = + 2$

Vertex$\to \left(x , y\right) = \left(2 , 2\right)$
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$\textcolor{b l u e}{\text{Determine the y-intercept}}$

Consider the given equation:$\text{ } = - {x}^{2} + 4 x \textcolor{m a \ge n t a}{- 2}$

${y}_{\text{intercept}} = \textcolor{m a \ge n t a}{- 2}$

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$\textcolor{b l u e}{\text{Determine the x-intercept}}$

Using $y = a {x}^{2} + b x + c \text{ where } x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$a = - 1 \text{; "b=+4"; } c = - 2$

$x = \frac{- 4 \pm \sqrt{{\left(4\right)}^{2} - 4 \left(- 1\right) \left(- 2\right)}}{2 \left(- 1\right)}$

I will let you finish this bit.