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salantis [7]
3 years ago
11

How to write y= 2x^2 + 8x +7 in standard (vertex) form

Mathematics
1 answer:
Anni [7]3 years ago
3 0
2x^2+8x+7= \\
2(x^2+4x)+7= \\
2(x^2+4x+4-4)+7= \\
2((x+2)^2-4)+7= \\
2(x+2)^2-8+7= \\
2(x+2)^2-1 \\ \\
\boxed{y=2(x+2)^2-1}
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Answer:

y=5x+4

Step-by-step explanation:

First get slope:

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The y-intercept is where the graph hits the y-axis. That is also the point on the graph where x = 0. We are already given that, with point (0,4).

The y-intercept is 4.

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