How do I find the x-intercepts of a quadratic function in vertex form ${(x + 4.5)}^{2} - 6.25$ $(x+4.5)^2-6.25$ ?

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Answer 1 · 2015-05-24T13:04:10

For a quadratic equation of the form
$y =$ $y =$ and expression in $x$ $x$

The x-intercepts occur at the points where $y = 0$ $y=0$

Given ( $y =$ $y=$ ) $({(x + 4.5)}^{2} - 6.25)$ $((x+4.5)^2-6.25)$
We need to solve
${(x + 4.5)}^{2} - 6.25 = 0$ $(x+4.5)^2 -6.25 = 0$

${(x + 4.5)}^{2} = 6.25$ $(x+4.5)^2 = 6.25$

$x + 4.5 = \pm 2.5$ $x+4.5 = +- 2.5$

$x = - 4.5 \pm 2.5$ $x = -4.5 +- 2.5$

The x-intercepts are
$(- 2)$ $(-2)$ and $(- 7)$ $(-7)$

How do I find the x-intercepts of a quadratic function in vertex form (x+4.5)^2-6.25(x+4.5)2−6.25(x+4.5)^2-6.25?