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.