# How do you graph the quadratic function and identify the vertex and axis of symmetry for y=1/2x^2+4x+5?

Jul 17, 2017

x_("intercpts")=-4+-sqrt(6)larr" Exact value"
${y}_{\text{intercept}} = 5$
Vertex $\to \left(x , y\right) = \left(- 4 , - 3\right)$

#### Explanation:

color(blue)("The different way to find "x_("vertex"))

Write as: $y = \frac{1}{2} \left({x}^{2} + 8 x\right) + 5$

Axis of symmetry $\to {x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times 8 = - 4$

${y}_{\text{vertex}} = \frac{1}{2} {\left(- 4\right)}^{2} + 4 \left(- 4\right) + 5 = - 3$

Vertex$\to \left(x , y\right) = \left(- 4 , - 3\right)$
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$\textcolor{b l u e}{\text{The formula method for x-intercepts}}$

Given: $y = \frac{1}{2} {x}^{2} + 4 x + 5$

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

This really is worth committing to memory.

Where $a = \frac{1}{2} \text{; "b=4"; } c = 5$

$x = \frac{- 4 \pm \sqrt{{4}^{2} - 4 \left(\frac{1}{2}\right) \left(5\right)}}{2 \left(\frac{1}{2}\right)}$

$x = - 4 \pm \sqrt{6} \leftarrow \text{ Exact value}$

$x \approx - 1.55 \mathmr{and} - 6.45$ to 2 decimal places
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$\textcolor{b l u e}{\text{Determine the vertex}}$

${x}_{\text{vertex}}$ is mid point of intercepts which is:

$\frac{\left(- 4 - \sqrt{6}\right) + \left(- 4 + \sqrt{6}\right)}{2} = - 4$

by substitution ${y}_{\text{vertex}} = - 3$
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$\textcolor{b l u e}{\text{Determine the y-intercept}}$

It is the value of the constant $c = 5$