# How do you find the x and y intercepts for f(x) = -x^2 + 6x + 8?

Apr 26, 2017

${x}_{\text{intercepts}} \to - 2 \mathmr{and} - 4$

${y}_{\text{intercept}} = + 8$

#### Explanation:

Given:$\text{ } f \left(x\right) = - {x}^{2} + 6 x + 8$

The use of the term $f \left(x\right)$ is that of giving a label (name) to a particular mathematical construct. The $f$ is the name and the $\left(x\right)$ bit means it is applied to $x$

So $f \left(a\right)$ would mean: $- {a}^{2} + 6 a + 8$
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$\textcolor{b l u e}{\text{Determine x-intercepts}}$

Observe that $2 \times 4 = 8 \text{ and that } 2 + 4 = 6$

so we can write: $y = \left(x + 2\right) \left(x + 4\right)$

The x-intercept is at $y = 0$ so :

Set $y = 0 = \left(x + 2\right) \left(x + 4\right)$

Thus for this condition $x = - 2 \mathmr{and} x = - 4$
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$\textcolor{b l u e}{\text{Determine y-intercept}}$

When the equation is in form $y = a {x}^{2} + b x + c$ then

${y}_{\text{intercept")=c=+8 color(red)(larr" Shortcut}}$

You obtain this value by substituting $0 \text{ for } x$