Question

date: solving quadratics day 3 5. solve the following function by graphing $-2x^2 + 4x + 6 = 0$

Explanation:

Step1: Rewrite in vertex form

First, factor out -2 from the first two terms:
$$-2(x^2 - 2x) + 6 = 0$$
Complete the square inside the parentheses:
$$-2(x^2 - 2x + 1 - 1) + 6 = 0$$
$$-2((x-1)^2 - 1) + 6 = 0$$
$$-2(x-1)^2 + 2 + 6 = 0$$
$$y = -2(x-1)^2 + 8$$

Step2: Find vertex and intercepts

Vertex is at $(1, 8)$ (from vertex form $y=a(x-h)^2+k$ where $(h,k)$ is vertex).
Find y-intercept: set $x=0$
$$y = -2(0)^2 + 4(0) + 6 = 6$$
So y-intercept is $(0,6)$.

Step3: Find x-intercepts (solutions)

Set $y=0$, solve $-2x^2 + 4x + 6 = 0$
Divide by -2:
$$x^2 - 2x - 3 = 0$$
Factor:
$$(x-3)(x+1) = 0$$
So $x=3$ and $x=-1$. These are the x-intercepts, which are the solutions to the equation.

Step4: Verify via graph

Plot vertex $(1,8)$, y-intercept $(0,6)$, and x-intercepts $(-1,0)$ and $(3,0)$, then draw the downward-opening parabola (since $a=-2<0$). The parabola crosses the x-axis at $x=-1$ and $x=3$, confirming the solutions.

Answer:

$x = -1$ and $x = 3$