Plz help with this question.If can't do it don't answer.
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Answer:
Step-by-step explanation:
We have to graph a line y = 3 - x which has the slope = -1 and y intercept 3.
We will select two points where line intersects at x = 0 and y = 0
The given line will intersect x-axis at (3, 0) and at y- axis (0, 3).
Joining these two points we can draw a straight line showing y = -x + 3
Now we will draw the parabola given by equation y = x² + x - 12
We will convert this equation in vertex form first to get the vertex and line of symmetry.
Standard equation of a parabola in vertex form is
y = (x - h)² + k
Where (h, k) is the vertex and x = h is the line of symmetry.
y = x² + x - 12
y = x² + 2(0.5)x + (0.5)²- (0.5)²-12
y = (x + 0.5)² - 12.25
Therefore, vertex will be (-0.5, -12.25) and line of symmetry will be x = 0.5
For x intercept,
0 = (x + 0.5)² - 12.25
x + 0.5 = ±√12.25
x = -0.5 ± 3.5
x = -4, 3
For y- intercept,
y = (0+0.5)² - 12.25
= 0.25 - 12.25
y = -12
So the parabola has vertex (-0.5, - 12.25), line of symmetry x = 0.5, x intercept (4, 0), (and y-intercept (0, -12).
Now we have to find the points of intersection of the given line and parabola.
For this we will replace the values of y
3 - x = x² + x - 12
x² + 2x - 15 = 0
x² + 5x - 3x - 15 = 0
x(x + 5) - 3(x + 5) = 0
(x - 3)(x + 5) = 0
x = 3, -5
For x = 3
y = 3- 3 = 0
For x = -5
y = 3 + 5 = 8
Therefore, points of intersection will be (3, 0) and (-5, 8)
