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.

Ay = x^2 − 6x − 7

x − y = 1

By = x^2 − 6x + 7

x + y = −1

Cy = x^2 + 6x − 7

x − y = 1

Dy = x^2 + 6x + 7

x + y = 1

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 class=

Verified Answer :

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:

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.