How do you find the vertex and intercepts for $y = (x + 5)^2 – 2$ ?

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Answer 1 · 2016-07-19T10:00:04

Vertex->(x,y)=(-5,-2)#

$y_("intercept")->y=23$

$x_("intercept")$ as follows:
$x=-5+-sqrt(2)$

$x~~-6.414$ to 3 decimal places
$x~~-3.586$ to 3 decimal places

$color(blue)("Determine the vertex")$

This is already in a vertex form the of quadratic equation so you can read off the coordinates of the vertex directly.

$x_("vertex")=(-1)xx(+5) = -5$
$y_("vertex")=-2$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$color(blue)("Determine y intercept")$

y intercept is at $x=0$

$=>y=(0+5)^2-2 -> y=23$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$color(blue)("Determine x intercept")$

Set $y=0$ giving

$0=(x+5)^2-2$

Add 2 to both sides

$0+2=(x+5)^2-2+2$

$=>2=(x+5)^2" "vec("swap round") " " (x+5)^2=2$

Square root both sides

$x+5=sqrt(2)$

Subtract 5 from both sides

$x=-5+-sqrt(2)$

$x~~-6.414$ to 3 decimal places
$x~~-3.586$ to 3 decimal places

Tony B

How do you find the vertex and intercepts for y = (x + 5)^2 – 2y = (x + 5)^2 – 2?