How do you find the x and y intercepts for $y = {(x - 8)}^{2} - 4$ $y=(x-8)^2-4$ ?

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How do you find the x and y intercepts for $y = {(x - 8)}^{2} - 4$ $y=(x-8)^2-4$ ?

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Answer 1 · 2016-12-06T11:56:18

For the $y$ $y$ axis just calculate $y$ $y$ for $x = 0$ $x=0$ .
And the same goes for the $x$ $x$ axis.

Explanation:

If you find yourself in the $y$ $y$ axis, you'll notice that there's no horizontal movement, which means that your $x$ $x$ is equal to $0$ $0$ .
Thus, if you want to find the $y$ $y$ axis interception you will have to impose the condition of $x = 0$ $x=0$ , this way you will have:

$x = 0 \to y = {(0 - 8)}^{2} - 4 = {(- 8)}^{2} - 4 = 64 - 4 = 60 \to$ $x=0 ->y=(0-8)^2-4=(-8)^2-4=64-4=60 ->$ Interception point $(0, 60)$ $(0, 60)$

And the same reasoning is applied for the $x$ $x$ axis. No vertical movement means $y = 0$ $y=0$ , in this case you will have to solve the complete second degree equation:

${(x - 8)}^{2} - 4 = 0 \to x^{2} - 2 \cdot 8 \cdot x + 64 - 4 = 0 \to x^{2} - 16 x + 60 = 0 \to x = \frac{16 \pm \sqrt{{(- 16)}^{2} - 4 \cdot 1 \cdot 60}}{2} \to x = \frac{16 \pm \sqrt{16}}{2} = \frac{16 \pm 4}{2}$ $(x-8)^2-4=0 -> x^2-2*8*x+64-4=0 -> x^2-16x+60=0 -> x=(16+-sqrt((-16)^2-4*1*60))/2 -> x=(16+-sqrt(16))/2=(16+-4)/2$

This gives us two interception points for the $x$ $x$ axis:

$x = \frac{16 + 4}{2} = 10$ $x=(16+4)/2=10$ and $x = \frac{16 - 4}{2} = 6$ $x=(16-4)/2=6$

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How do you find the x and y intercepts for $y = {(x - 8)}^{2} - 4$ $y=(x-8)^2-4$ ?

1 Answer

Explanation:

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How do you find the x and y intercepts for y=(x−8)2−4y=(x-8)^2-4?

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How do you find the x and y intercepts for $y = {(x - 8)}^{2} - 4$ $y=(x-8)^2-4$ ?