Question

question 7 (multiple choice worth 6 points) (06.04r mc)
which graph correctly solves the equation below?
$2x^2 - 3 = -x^2$

Explanation:

Step1: Rearrange the given equation

First, rewrite the equation to find the functions to graph, and also solve for the intersection points.
Start with $2x^2 - 3 = -x^2$
Bring all terms to one side: $2x^2 + x^2 - 3 = 0$

Step2: Simplify to find intersection points

Combine like terms: $3x^2 - 3 = 0$
Factor and solve:
$3(x^2 - 1) = 0$
$x^2 - 1 = 0$
$(x-1)(x+1)=0$
So $x=1$ and $x=-1$. Substitute back to find $y$: use $y=-x^2$, so when $x=1$, $y=-1$; when $x=-1$, $y=-1$. The intersection points are $(-1, -1)$ and $(1, -1)$.

Step3: Identify the two functions

Rewrite the original equation as two separate functions to graph:
Left side: $y=2x^2 - 3$ (upward-opening parabola, vertex at $(0, -3)$)
Right side: $y=-x^2$ (downward-opening parabola, vertex at $(0, 0)$)

Step4: Match to the correct graph

We need a graph with an upward-opening parabola (vertex at (0,-3)) and a downward-opening parabola (vertex at (0,0)), intersecting at $(-1, -1)$ and $(1, -1)$.

Answer:

The first graph (topmost option, with upward-opening red parabola and downward-opening black parabola intersecting at (-1,-1) and (1,-1))