Question

solving quadratics day 3

1. solve the following function by graphing

$-2x^2 + 4x + 6 = 0$

1. solve the following function by graphing

$-x^2 - 6x - 5 = 0$

Explanation:

Problem 5: $-2x^2 + 4x + 6 = 0$

Step1: Find vertex (x-coord)

Use $x = -\frac{b}{2a}$. For $a=-2, b=4$:
$x = -\frac{4}{2(-2)} = 1$

Step2: Find vertex (y-coord)

Substitute $x=1$ into equation:
$y = -2(1)^2 + 4(1) + 6 = -2 + 4 + 6 = 8$
Vertex: $(1, 8)$

Step3: Find x-intercepts (roots)

Factor the quadratic:
$-2x^2 + 4x + 6 = 0 \implies -2(x^2 - 2x - 3) = 0 \implies -2(x-3)(x+1) = 0$
Set factors to 0: $x-3=0$ or $x+1=0$

Step4: Graph key points

Plot vertex $(1,8)$, x-intercepts $(-1,0)$ and $(3,0)$, then draw downward-opening parabola.

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Problem 6: $-x^2 - 6x - 5 = 0$

Step1: Find vertex (x-coord)

Use $x = -\frac{b}{2a}$. For $a=-1, b=-6$:
$x = -\frac{-6}{2(-1)} = -3$

Step2: Find vertex (y-coord)

Substitute $x=-3$ into equation:
$y = -(-3)^2 -6(-3) -5 = -9 + 18 -5 = 4$
Vertex: $(-3, 4)$

Step3: Find x-intercepts (roots)

Factor the quadratic:
$-x^2 -6x -5 = 0 \implies -(x^2 +6x +5) = 0 \implies -(x+5)(x+1) = 0$
Set factors to 0: $x+5=0$ or $x+1=0$

Step4: Graph key points

Plot vertex $(-3,4)$, x-intercepts $(-5,0)$ and $(-1,0)$, then draw downward-opening parabola.

Answer:

1. For $-2x^2 + 4x + 6 = 0$, the solutions (x-intercepts) are $x=-1$ and $x=3$.
2. For $-x^2 - 6x - 5 = 0$, the solutions (x-intercepts) are $x=-5$ and $x=-1$.