Question

Which equation does the graph of the systems of equations solve?

ANSWER CHOICES:
x2 − 2x + 8 = x2 − 8x − 16
x2 + 6x + 8 = −x2 − 8x − 16
−x2 − 2x + 8 = 2x2 − 8x − 16
−x2 − 2x + 8 = 2x2 − 8x + 16

Answer 1

@princeharryyy
would you mind helping me, please?

Answer 2

just hold on a little bit. I will.

Answer 3

well, you could look for the axis of symmetry and that will give you the b and the a values for your equations.

Answer 4

well it is B again.

Answer 5

just insert the points in option 2 both of them satisfies.

Answer 6

So, it is B

Answer 7

You have two parabolas. A parabola with 2 real, distinct roots crosses the x-axis twice, once at each root. You have a parabola that crosses at \(x=-2\) and \(x=-4\) so those two values of \(x\) will both make the value of the equation be equal to \(0\). And you can also write the polynomial that has those roots in the form \[P(x) = a(x-r_1)(x-r_2)\]where \(a\) is a constant (often \(1\)) and \(r_1, r_2\) are the roots.

If we have roots of \(-2, -4\) as determined from looking at the graph, that means the parabola has the form\[P(x) = a(x-(-2))(x-(-4)) = a(x+2)(x+4)\]If we expand that factors, that becomes \[P(x) = a(x*2 + x*2 + x*4 + 2*4) = a(x^2 +6x +8)\]

Any of the answer choices that does not have \(x^2+6x+8\) or a multiple of it on one side or the other must not be a valid answer choice. That immediately rules out all but choice B.

Answer 8

A parabola which touches the x-axis only in one place has a double root at that point, and can be written \[P(x) = a(x-r_1)^2\]Our intersecting parabola has its contact with the x axis at \(x=-4\), so it is of the form \[P(x) = a(x-(-4))^2 = a(x+4)^2 = a(x*x+4*x+4*x+4*4)\]\[P(x) = a(x^2+8x+16)\]and if we look at option B, we see that the right hand side is \[-x^2-8x-16\]which can be written as \[(-1)x^2 + (-1)(8x) + (-1)(16)=(-1)(x^2+8x+16)\]which means it is in the necessary form for a parabola which only touches the x axis at \(x=-4\).