Question

section 3.5 homework

1. solve the system by graphing.

$y = -x^2 - 2$
$y = 4(x + 1) - 3$

a. $\square$ $(-1, -1)$
b. $\square$ $(-1, -3)$
c. $\square$ $(1, 1)$
d. $\square$ $(-7, 9)$
e. $\square$ $(-3, -11)$
f. $\square$ $(-5, -9)$
g. $\square$ $\left(-\frac{3}{5}, \frac{1}{2}\
ight)$
h. $\square$ no real solution

Explanation:

Step1: Set the equations equal

Since \( y = -x^2 - 2 \) and \( y = 4(x + 1) - 3 \), we set them equal: \(-x^2 - 2 = 4(x + 1) - 3\)

Step2: Simplify the equation

Expand the right side: \(-x^2 - 2 = 4x + 4 - 3\) → \(-x^2 - 2 = 4x + 1\)
Move all terms to one side: \(-x^2 - 4x - 3 = 0\) → Multiply by -1: \(x^2 + 4x + 3 = 0\)

Step3: Factor the quadratic

Factor \(x^2 + 4x + 3\): \((x + 1)(x + 3) = 0\)

Step4: Solve for x

Set each factor to zero: \(x + 1 = 0\) → \(x = -1\); \(x + 3 = 0\) → \(x = -3\)

Step5: Find corresponding y-values

For \(x = -1\): \(y = -(-1)^2 - 2 = -1 - 2 = -3\) → Point \((-1, -3)\)
For \(x = -3\): \(y = -(-3)^2 - 2 = -9 - 2 = -11\) → Point \((-3, -11)\)

Answer:

b. \((-1, -3)\), e. \((-3, -11)\)