Use graphing to find the solutions to the system of equations.
y=x^2+6x+8
y=x+4
(1 point)
Responses
graph a- The line and the quadratic function both pass through the points left-parenthesis negative 4 comma 0 right-parenthesis and left-parenthesis negative 1 comma negative 3 right-parenthesis. The quadratic function has a maximum at left-parenthesis negative 3 comma 1 right-parenthesis.
(—4, 0)
(—1, —3)
graph b- The line and the quadratic function both pass through the points left-parenthesis negative 3 comma 0 right-parenthesis and left-parenthesis negative 2 comma 1 right-parenthesis. The quadratic function has a minimum at left-parenthesis negative 3 comma 0 right-parenthesis and also passes through the point left-parenthesis negative 4 comma 1 right-parenthesis.
(—3, 0)
(—2, 1)
graph c- The line and the quadratic function both pass through the points left-parenthesis negative 4 comma 0 right-parenthesis and left-parenthesis negative 1 comma 3 right-parenthesis. The quadratic function has a minimum at left-parenthesis negative 3 comma negative 1 right-parenthesis.
(—4, 0)
(—1, 3)
graph d- The line and the quadratic function both pass through the points left-parenthesis negative 3 comma 0 right-parenthesis and (- 2, - 1). The quadratic function has a maximum at (-3, 0) and also passes through the point (-4, -1)
(—3, 0)
(—2, —1)
The correct answer is graph d- The line and the quadratic function both pass through the points left-parenthesis negative 3 comma 0 right-parenthesis and left-parenthesis negative 2 comma negative 1 right-parenthesis. The quadratic function has a maximum at left-parenthesis negative 3 comma 0 right-parenthesis and also passes through the point left-parenthesis negative 4 comma negative 1 right-parenthesis.
(-3, 0)
(-2, -1)
To find the solutions to the system of equations y = x^2 + 6x + 8 and y = x + 4 using graphing, we can plot the graphs of both equations and find the points where they intersect.
To graph the first equation y = x^2 + 6x + 8, we can start by finding a few points on the graph. We can substitute different values for x and calculate the corresponding y values.
Let's choose x = -4:
y = (-4)^2 + 6(-4) + 8
y = 16 - 24 + 8
y = 0 - 8
y = -8
So one point on the graph is (-4, 0).
Now let's choose x = -1:
y = (-1)^2 + 6(-1) + 8
y = 1 - 6 + 8
y = 3
So another point on the graph is (-1, 3).
For the second equation y = x + 4, we can simply use the slope-intercept form to graph it. The slope is 1 and the y-intercept is 4, so the line passes through the point (0, 4).
Now let's plot these points and draw the graphs:
The quadratic function graph passes through the points (-4, 0) and (-1, 3), and it has a maximum at (-3, 1).
The line graph passes through the point (-4, 0) and (0, 4).
The solution to the system of equations is the point where the graphs intersect, which is approximately (-3, 0).
So the answer is graph d: (-3, 0) and (-2, -1).
I hope this helps!
To find the solutions to the system of equations y = x^2 + 6x + 8 and y = x + 4 using graphing, we need to plot the graphs of both equations on the same coordinate plane and find the points where they intersect.
To plot the graph of y = x^2 + 6x + 8, we can start by finding a few points on the graph. One way to do this is by plugging in a few x-values and calculating the corresponding y-values. For example, when x = -4, we have:
y = (-4)^2 + 6(-4) + 8
y = 16 - 24 + 8
y = 0
So we have one point on the graph: (-4, 0).
Similarly, when x = -1, we have:
y = (-1)^2 + 6(-1) + 8
y = 1 - 6 + 8
y = 3
So another point on the graph is (-1, 3).
These two points are enough to give us an idea of the shape of the quadratic function y = x^2 + 6x + 8. We can now sketch this graph passing through the points (-4, 0) and (-1, 3).
Next, let's consider the equation y = x + 4. This equation represents a straight line. We can find a couple of points on this line by plugging in x-values and calculating the corresponding y-values. For example, when x = -4, we have:
y = -4 + 4
y = 0
So one point on the line is (-4, 0).
Similarly, when x = -3, we have:
y = -3 + 4
y = 1
So another point on the line is (-3, 1).
Now, we can sketch the line passing through the points (-4, 0) and (-3, 1).
The solution to the system of equations is the point(s) where the line and the quadratic function intersect on the graph. By examining the options provided, we can see that the correct graph is option d. The line and the quadratic function intersect at (-3, 0) and (-2, -1), which are the solutions to the system of equations.