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

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Answer 1 · 2016-04-26T15:26:45

No x intercepts
$y_("intercept")= -15/6$

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

The y intercept is at $x=0$

$=>y_("intercept")=-0.3(0+2)^2-2$

$y_("intercept")=(4xx0.3)-2 = (4xx3/10)-2$
$color(blue)(y_("intercept")= -(4xx3/10)-2 = -15/5)$
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The x intercepts are at $y=0$

$=>0=-0.3(x+2)^2-2$

Add 2 to both sides

$0+2=-0.3(x+2)^2-2+2$

$2=-0.3(x+2)^2$

Divide both sides by $0.3$

$2/0.3 =-0.3/0.3(x+2)^2$

But $-0.3/0.3=-1$

$2/0.3=-(x+2)^2$

Multiply both sides by $-1$

$-2/0.3=+(x+2)^2$

Square root both sides

$x+2=sqrt( -2/0.3)$

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

Tony B

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