Question #c34a0

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Question #c34a0

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Answer 1 · 2017-11-20T07:49:46

$y = \frac{1}{2} {(x + 5)}^{2} + 1$ $y=1/2(x+5)^2+1$
or
$y = \frac{1}{2} x^{2} + 5 x + 13.5$ $y=1/2x^2+5x+13.5$

Explanation:

It's not 1-across, 1-up so the coefficient isn't one. It's 2 across, 2 up. Hence $y = k {(x \pm b)}^{2} + c$ $y=k(x±b)^2 + c$ , $k = \frac{1}{2}$ $k=1/2$

So now we have $y = \frac{1}{2} {(x \pm b)}^{2} + c$ $y=1/2(x±b)^2+c$

b and c are much easier to find - they're just how much it's shifted. The vertex has been shifted horizontally 5 to the left and vertically 1 up so $b = 5$ $b=5$ and $c = 1$ $c=1$

Therefore $y = \frac{1}{2} {(x + 5)}^{2} + 1$ $y=1/2(x+5)^2+1$ is your answer.

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Question #c34a0

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