Question

Use graphing technology to find the solutions to the linear-quadratic system of equations:

y=x2−3x−4

y=−x−1

Which answer gives the solutions to the system of equations?

(1 point)
Responses

(−1,0) and (4,0)
open paren negative 1 comma 0 close paren and open paren 4 comma 0 close paren

(0,−1) and (−4,3)
open paren 0 comma negative 1 close paren and open paren negative 4 comma 3 close paren

(−1,0) and (3,−4)
open paren negative 1 comma 0 close paren and open paren 3 comma negative 4 close paren

(0,−1) and (0,−4)

Answer 1

To find the solutions to the system of equations represented by:

1. \( y = x^2 - 3x - 4 \) (a quadratic equation)
2. \( y = -x - 1 \) (a linear equation)

we need to determine the points at which these two curves intersect.

To do this algebraically, we can set the two equations equal to each other:

\[
x^2 - 3x - 4 = -x - 1
\]

Next, we rearrange the equation to set it to 0:

\[
x^2 - 3x + x - 4 + 1 = 0
\]

Simplifying gives:

\[
x^2 - 2x - 3 = 0
\]

Now we can factor the quadratic:

\[
(x - 3)(x + 1) = 0
\]

Setting each factor to zero gives us:

1. \( x - 3 = 0 \) → \( x = 3 \)
2. \( x + 1 = 0 \) → \( x = -1 \)

Now, we substitute these \( x \)-values back into either original equation to find the corresponding \( y \)-values. Let's use the linear equation \( y = -x - 1 \).

For \( x = 3 \):

\[
y = -3 - 1 = -4
\]
→ The point is \( (3, -4) \).

For \( x = -1 \):

\[
y = -(-1) - 1 = 0
\]
→ The point is \( (-1, 0) \).

Thus, the solutions to the system of equations are the points \( (-1, 0) \) and \( (3, -4) \).

The correct answer is:

**(−1,0) and (3,−4)**
open paren negative 1 comma 0 close paren and open paren 3 comma negative 4 close paren.