Question

graph the system below. then write its solution. \

$$\begin{cases} y = -2x^2 + 3 \\\\ y = -2x - 1 \\end{cases}$$

if there is more than one solution, use the \or\ button.

Explanation:

Step1: Set equations equal

$-2x^2 + 3 = -2x - 1$

Step2: Rearrange to standard quadratic

$-2x^2 + 2x + 4 = 0$
Divide by $-2$: $x^2 - x - 2 = 0$

Step3: Factor quadratic

$(x - 2)(x + 1) = 0$

Step4: Solve for $x$

$x - 2 = 0 \implies x=2$; $x + 1 = 0 \implies x=-1$

Step5: Find corresponding $y$

For $x=2$: $y=-2(2)-1=-5$
For $x=-1$: $y=-2(-1)-1=1$

Answer:

The solutions are $(2, -5)$ or $(-1, 1)$

(For graphing: Plot the downward-opening parabola $y=-2x^2+3$ with vertex at $(0,3)$, and the line $y=-2x-1$ with y-intercept $(0,-1)$ and slope $-2$. The two intersection points are the solutions above.)