# How do you find the vertex and intercepts for f(x)=3 -8x -4x^2?

Mar 26, 2016

y_("intercept") = "the constant "=3

x_("intercpts") -> x~~0.3229" or "x~~ -2.3229

$\text{Vertex "->" } \left(x , y\right) = \left(- 1 , 7\right)$

#### Explanation:

Given:$\text{ } y = - 4 {x}^{2} - 8 x + 3$

This equation does not have whole numbers as roots so the formula has to be used to solve for ${x}_{\text{intercepts}}$.

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

color(green)(y_("intercept") = "the constant "=3)

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$\textcolor{b l u e}{\text{Determine the vertex}}$

Write the equation as:
$\text{ } y = - 4 \left({x}^{2} + 2 x\right) + 3 \to$part way to vertex form

Consider $2 \text{ from } 2 x$

Apply $\left(- \frac{1}{2}\right) \times 2 = - 1$

$\textcolor{b l u e}{{x}_{\text{vertex}} = - 1}$

By substitution:

$\textcolor{g r e e n}{{y}_{\text{vertex}} = - 4 {\left(- 1\right)}^{2} - 8 \left(- 1\right) + 3 = + 7}$
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color(blue)("Determine "x_("intercepts"))

Using: $y = a {x}^{2} + b x + c$ where $\text{ "a=-4"; " b=-8"; } c = 3$

$\text{ and } x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$x = \frac{+ 8 \pm \sqrt{{\left(- 8\right)}^{2} - 4 \left(- 4\right) \left(3\right)}}{2 \left(- 4\right)}$

$x = \frac{+ 8 \pm \sqrt{112}}{- 8}$

$x = \frac{8 \pm \sqrt{{2}^{2} \times 28}}{-} 8$

$x = \frac{8 \pm 2 \sqrt{28}}{-} 8$

$\textcolor{g r e e n}{x = - 1 \pm \frac{\sqrt{28}}{4} \approx 0.3229 \text{ or } x \approx - 2.3229}$

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