Question

675 x 576

Answer 1

Equation:
\[y - 2 = 2(x + 1)^2\]
Expand the brackets
\[y - 2 = 2(x^2+2x+ 1)\]
Expand the brackets again
\[y - 2 = 2x^2+4x+ 2\]
Move the \(2\) to the RHS
\[y=2x^2+4x+4\]
Axis of symmetry is:
\[\frac{-b}{2a}\]
In your equation:
\(a\) = 2
\(b\) = 4
Sub in the values:
\[\frac{-b}{2a}\]\[\frac{-1*4}{2*4}\]\[\frac{-4}{8}\]\[-0.5\]
Therefore, the axis of symatry is \(0.5\)

Answer 2

Hope that helps, @KierseyClemons

Answer 3

Sorry, it meant to be negative @KierseyClemons, just forgot to type it ;)

Answer 4

To find the \(x\) intercept, you need to use the quadratic formula:
\[\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
In your equation:
\(a=2\)
\(b=4\)
\(c=4\)

Simply sub the values into the equation to get the answer
\[\frac{-1*4\pm\sqrt{4^2-1*(4*2*4)}}{2*2}\]

Answer 5

Have you learned how to find the \(x-intercept\)?

Answer 6

Then why do you need to find it @KierseyClemons? Does it ask for it?

Answer 7

I just wanted to know the x intercept. Don't worry about it lol

Answer 8

Oh, thats fine. You would use the equation:
\[\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
And change \(a, b, c\) to your values

Answer 9

and how would that look? like I was confused on it and wanted to better myself

Answer 10

The answer @KierseyClemons? Or the steps?

Answer 11

Hmm. I'll try

Answer 12

\[\frac{−b\pm\sqrt{b^2−4ac}}{2a}\]

In your equation:
\(a=2\)
\(b=4\)
\(c=4\)

Simply sub the values into the equation to get the answer
\[\frac{−4\pm\sqrt{4^2−4*2*4}}{2*2}\]
\[\frac{−4\pm\sqrt{16-32}}{4}\]
\[\frac{−4\pm\sqrt{-16}}{4}\]

Now, can you see that the number inside the \(\sqrt{}\) is negative? That means there is no \(x-intercept\).

Therefore, there is no \(x-intercept\) for that equation. Hope that helps @KierseyClemons

Answer 13

No problem @KierseyClemons :) Probably a bad example, cause there is no \(x-intercept\) in that equation ;)