Question

Which equation could be solved using the graph above?
a. x^2+x-2=0
b. x^2+2x+1=0
c. x^2-1=0
d. x^2-2x+1=0

Answer 1

Your vertex is (0,-1).

Answer 2

For example,
\(\color{#000000 }{ \displaystyle 0=x^2+x-2 }\)

is the equation that solves for y-intercepts
(when y=0), of,
\(\color{#000000 }{ \displaystyle y=x^2+x-2 }\).

Wen I solve for vertex I get,

\(\color{#000000 }{ \displaystyle y=\left(x^2+x+\frac{1}{4}\right)-2-\frac{1}{4} }\)

\(\color{#000000 }{ \displaystyle y=\left(x^2+2\cdot\frac{1}{2}\cdot x+\left(\frac{1}{2}\right)^2\right)-2-\frac{1}{4} }\)

\(\color{#000000 }{ \displaystyle y=\left(x+\frac{1}{2} \right)^2-2-\frac{1}{4} }\)

Answer 3

So, option a does not have (0,-1) as the vertex

Answer 4

You just need to solve other options for the vertex, by completing the square.

Answer 5

Or, by graphing.

Answer 6

The graph shows the curve crossing the x-axis at \(x=-1\), and \(x = 1\). Which of the equations in the answer choices gives a value of \(0\) when evaluated at \(x = -1\) and \(x=1\)?

When you see a symmetrical answer like that, you should immediately think "difference of squares"...