Question

Graph the line and the parabola and find the points of intersection: y= 3-x ; y=x(squared)+x-12

Answer 1

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)