# How do you find the intercept and vertex of y= - .3(x+2)^2-2?

Apr 26, 2016

No x intercepts
${y}_{\text{intercept}} = - \frac{15}{6}$

#### Explanation:

Assumption: The question has -.3 which is assumed to mean - 0.3

The y intercept is at $x = 0$

$\implies {y}_{\text{intercept}} = - 0.3 {\left(0 + 2\right)}^{2} - 2$

${y}_{\text{intercept}} = \left(4 \times 0.3\right) - 2 = \left(4 \times \frac{3}{10}\right) - 2$
$\textcolor{b l u e}{{y}_{\text{intercept}} = - \left(4 \times \frac{3}{10}\right) - 2 = - \frac{15}{5}}$
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The x intercepts are at $y = 0$

$\implies 0 = - 0.3 {\left(x + 2\right)}^{2} - 2$

$0 + 2 = - 0.3 {\left(x + 2\right)}^{2} - 2 + 2$

$2 = - 0.3 {\left(x + 2\right)}^{2}$

Divide both sides by $0.3$

$\frac{2}{0.3} = - \frac{0.3}{0.3} {\left(x + 2\right)}^{2}$

But $- \frac{0.3}{0.3} = - 1$

$\frac{2}{0.3} = - {\left(x + 2\right)}^{2}$

Multiply both sides by $- 1$

$- \frac{2}{0.3} = + {\left(x + 2\right)}^{2}$

Square root both sides

$x + 2 = \sqrt{- \frac{2}{0.3}}$

As we are trying to take the square root of a negative number it means that the graph does not cross the x-axis. So there are no x intercepts 