a) Given x + y = 2 and y = x² - x - 2, find the x-intercepts and y-intercept of the

parabola.
b) Find the turning point of the parabola.
c) Find the x intercept and y intercept of the line graph.
d) Find the point(s) of intersection of the line graph and the parabola.
e) Represent these equations on a graph, including all of the points determined in a-d
above. Be sure to fully label your graphs.

a) To find the x-intercepts of the parabola, we need to solve the equation y = x² - x - 2 for x when y equals 0.

Setting y equal to 0:
0 = x² - x - 2

To solve this quadratic equation, we can factor or use the quadratic formula. Let's use the quadratic formula:

x = (-b ± √(b² - 4ac)) / (2a)

In this case:
a = 1
b = -1
c = -2

Plugging the values into the formula, we have:
x = (-(-1) ± √((-1)² - 4(1)(-2))) / (2(1))

Simplifying:
x = (1 ± √(1 + 8)) / 2

x = (1 ± √9) / 2

x = (1 ± 3) / 2

So the x-intercepts are:
x₁ = (1 + 3) / 2 = 2
x₂ = (1 - 3) / 2 = -1

To find the y-intercept, we can substitute x = 0 into the equation y = x² - x - 2:

y = (0)² - (0) - 2
y = -2

Therefore, the y-intercept is -2.

b) The turning point of a parabola is the vertex. To find the turning point, we can use the formula x = -b / (2a).

In this case:
a = 1
b = -1

x = -(-1) / (2(1))
x = 1/2

To find the y-coordinate of the turning point, we substitute x = 1/2 into the equation y = x² - x - 2:

y = (1/2)² - (1/2) - 2
y = 1/4 - 1/2 - 2
y = -9/4

Therefore, the turning point is (1/2, -9/4).

c) To find the x-intercept and y-intercept of a line, we need its equation. However, we have not been given an equation for the line in this question. Please provide the equation for the line.

d) To find the point(s) of intersection of the line graph and the parabola, we need to solve the system of equations:

x + y = 2
y = x² - x - 2

Substituting the value of y from the second equation to the first equation:
x + x² - x - 2 = 2

Rearranging the terms:
x² - 2 = 2 - x

Bringing all the terms to one side:
x² + x -4 = 0

To solve this quadratic equation, we can factor or use the quadratic formula. Let's use the quadratic formula:

x = (-b ± √(b² - 4ac)) / (2a)

In this case:
a = 1
b = 1
c = -4

Plugging the values into the formula, we have:
x = (-1 ± √(1² - 4(1)(-4))) / (2(1))

Simplifying:
x = (-1 ± √(1 + 16)) / 2
x = (-1 ± √17) / 2

Therefore, the x-coordinate of the point(s) of intersection are:
x₁ = (-1 + √17) / 2
x₂ = (-1 - √17) / 2

To find the y-coordinate of the point(s) of intersection, we can substitute these x-values back into either the equation of the line or the equation of the parabola.

e) To represent these equations on a graph, you can plot the x-intercepts, y-intercept, turning point, and point(s) of intersection on a coordinate plane. Label each point with their respective coordinates. Draw the parabola and the line graph based on the given equations.

To find the x-intercepts and y-intercept of the parabola and line graph, as well as the turning point and points of intersection, we need to solve the system of equations formed by the given equations.

a) To find the x-intercepts of the parabola, we set y = 0 in the equation y = x² - x - 2. This gives us 0 = x² - x - 2. We can now factorize this equation or use the quadratic formula to find the values of x that make the equation equal to zero. Once we find the x-values, we can substitute them back into the equation y = x² - x - 2 to find the corresponding y-values. These x-intercepts represent the points where the parabola crosses the x-axis.

For the y-intercept of the parabola, we set x = 0 in the equation y = x² - x - 2. This gives us y = 0² - 0 - 2 = -2. So the y-intercept is the point (0, -2).

b) The turning point of a parabola can be found by completing the square or using the vertex formula. The vertex formula states that the x-coordinate of the turning point is given by x = -b/2a, where the equation of the parabola is in the form y = ax² + bx + c. In this case, our equation is y = x² - x - 2, so the x-coordinate of the turning point is x = -(-1)/(2 * 1) = 1/2. To find the y-coordinate of the turning point, substitute this x-value back into the equation y = x² - x - 2. Thus, the turning point is (1/2, -9/4).

c) To find the x-intercept of the line graph given by x + y = 2, we set y = 0 and solve for x. Therefore, x = 2. The x-intercept is the point (2, 0).

For the y-intercept, we set x = 0 and solve for y. Thus, 0 + y = 2, giving y = 2. The y-intercept is the point (0, 2).

d) To find the points of intersection between the line graph x + y = 2 and the parabola y = x² - x - 2, we can substitute one equation into the other and solve for x. Substituting y = x² - x - 2 into x + y = 2, we get x + (x² - x - 2) = 2. Simplifying this equation and solving for x will give us the x-coordinates of the points of intersection. Once we have the x-values, we can substitute them back into either equation to find the corresponding y-values.

e) To represent these equations on a graph, plot the x and y-intercepts, the turning point, and the points of intersection on a coordinate plane. Label the graph with the respective coordinates. Draw the line graph x + y = 2 as a straight line and the parabola y = x² - x - 2 as a curved line. Ensure that the graph is labeled accurately.