Question

for the equations $y = x + 1$ and $y = x^2 + 5x + 5$, what are the solutions?
○ a. $(-1, -5)$
○ b. no solution
○ c. $(6,16)$ and $(4,10)$
○ d. $(-2, -1)$

if you graph the system $y = x^2 - 4x + 3$ and $y = 2x - 1$, what will the graph show about the intersection points?
○ a. the line does not intersect the parabola
○ b. the line is tangent to the parabola
○ c. the line intersects the parabola at two points
○ d. the line intersects the parabola at only one point

when solving the system $y = x^2 - 6x + 5$ and $y = 5$ using substitution, what is the resulting quadratic equation after substituting $y = 5$?
○ a. $x^2 - 6x + 5 = 0$
○ b. $x^2 - 6x + 5 = 5$
○ c. $x^2 - 6x + 10 = 0$
○ d. $x^2 - 6x - 5 = 0$

Explanation:

First Question:

Step1: Set equations equal

Since \( y = x + 1 \) and \( y = x^2 + 5x + 5 \), substitute \( y \): \( x + 1 = x^2 + 5x + 5 \).

Step2: Rearrange to quadratic

\( x^2 + 4x + 4 = 0 \), factor: \( (x + 2)^2 = 0 \), so \( x = -2 \).

Step3: Find \( y \)

Substitute \( x = -2 \) into \( y = x + 1 \): \( y = -2 + 1 = -1 \). Solution is \( (-2, -1) \), which is option d.

Step1: Set equations equal

\( x^2 - 4x + 3 = 2x - 1 \).

Step2: Rearrange to quadratic

\( x^2 - 6x + 4 = 0 \).

Step3: Check discriminant

Discriminant \( D = (-6)^2 - 4(1)(4) = 36 - 16 = 20 > 0 \), so two real roots (two intersection points), option c.

Step1: Substitute \( y = 5 \)

Into \( y = x^2 - 6x + 5 \), get \( 5 = x^2 - 6x + 5 \).

Step2: Simplify

Subtract 5: \( x^2 - 6x + 5 - 5 = 0 \) → \( x^2 - 6x = 0 \)? Wait, no: original substitution is \( 5 = x^2 - 6x + 5 \), so \( x^2 - 6x + 5 = 5 \) (option b).

Answer:

d. (-2, -1)

Second Question: