Question

Q1. Which system of equations is represented by the graph?

a line is graphed through points negative 6 comma 7 and negative 1 comma 2. It intersects a parabola that opens up at these two points.

A. y = x^2 − 6x − 7
x − y = 1

B. y = x^2 − 6x + 7
x + y = −1

C. y = x^2 + 6x − 7
x − y = 1

D. y = x^2 + 6x + 7
x + y = 1

Answer 1

Let

$A(-6,7)\\B(-1,2)$

we know that

The points A and B must satisfy both equations

we proceed to verify each case

case A)

equation $1$

$y=x^{2} -6x-7$

check point A

for $x=-6$

y must be $7$

substitute

$y=(-6)^{2} -6*(-6)-7$

$y=36+36-7\\ y=65$

$65$ is not equal to $7$

so

The system of equations is not represented by the graph

It is not necessary to check point B

case B)

equation $1$

$y=x^{2} -6x+7$

check point A

for $x=-6$

y must be $7$

substitute

$y=(-6)^{2} -6*(-6)+7$

$y=36+36+7\\ y=79$

$79$ is not equal to $7$

so

The system of equations is not represented by the graph

It is not necessary to check point B

case C)

equation $1$

$y=x^{2} +6x-7$

check point A

for $x=-6$

y must be $7$

substitute

$y=(-6)^{2} +6*(-6)-7$

$y=36-36-7\\ y=-7$

$-7$ is not equal to $7$

so

The system of equations is not represented by the graph

It is not necessary to check point B

case D)

equation $1$

$y=x^{2} +6x+7$

check point A

for $x=-6$

y must be $7$

substitute

$y=(-6)^{2} +6*(-6)+7$

$y=36-36+7\\ y=7$

$7$ is equal to $7$------->is ok

check point B

for $x=-1$

y must be $2$

substitute

$y=(-1)^{2} +6*(-1)+7$

$y=1-6+7\\ y=2$

$2$ is equal to $2$----->is ok

equation $2$

$x+y=1$

check point A

for $x=-6$

y must be $7$

substitute

$-6+y=1\\ y=6+1\\ y=7$

$7$ is equal to $7$------->is ok

check point B

for $x=-1$

y must be $2$

substitute

$-1+y=1\\ y=1+1\\ y=2$

$2$ is equal to $2$----->is ok

therefore

the answer is the option D

$y=x^{2} +6x+7$

$x+y=1$

Answer 2

Answer:

The correct option is D.

Step-by-step explanation:

From the given graph it is noticed that the line and parabola intersect each other at points (-6,7) and (-1,2). It means each equation of the system must be satisfied by these points.

If a line passing through two points, then the equation of line is

$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

The equation of line is

$y-7=\frac{2-7}{-1-(-6)}(x-(-6))$

$y-7=-(x+6)$

$y-7=-x-6$

$y+x=1$                        .... (1)

Therefore the equation of line is $y+x=1$.

The standard form of the parabola is

$y=a(x-h)^2+k$

Where (h,k) is the vertex.

From the graph it is noticed that the vertex of the parabola is (-3,-2).

$y=a(x+3)^2-2$

The parabola passing through the point (-1,2).

$2=a(-1+3)^2-2$

$2=4a-2$

$a=1$

Therefore the equation of parabola is

$y=(1)(x+3)^2-2=x^2+6x+9-2=x^2+6x+7$   .... (2)

So, the system of equations contains equation (1) and (2). Option D is correct.