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3 years ago

Can someone help with this?

Mathematics

1 answer:

skelet666 [1.2K]3 years ago

6 0

Answer:

Step-by-step explanation:

only Ali's way would work in this situation. since there is a 5 on the right side, you can not use the zero product property. this only works if the right side is 0. so, you must use Ali's method.

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$y = {x}^{2} - 2$

Or if you want with the value of h too.

$y = {(x - 0)}^{2} - 2$

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$y = a {(x - h)}^{2} + k$

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From y = x²-2

$a = 1 \\ b = 0 \\ c = - 2$

Substitute these values in the formula.

$h = - \frac{0}{2(1)} \\ h = 0$

Therefore, h = 0.

$k = \frac{4(1)( - 2) - {0}^{2} }{4(1)} \\ k = \frac{ - 8}{4} \\ k = - 2$

Therefore, k = - 2.

From the vertex form, the vertex is at (h, k) = (0,-2). Substitute h = 0, a = 1 and k = -2 in the equation.

$y = a {(x - h)}^{2} + k \\ y = 1 {(x - 0)}^{2} - 2 \\ y = {(x)}^{2} - 2 \\ y = {x}^{2} - 2$

These type of equation where b = 0 can also be both standard and vertex form.

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