staright line y=x-1 meets curve
y=x^2-5x-8 at the points A and B.
the curve y=p+qx-2x^2 also passes through ponts A and B. what are the values of p and q??
helpppppplease...thanks
First, find the intersection of y = x-1 with y = x^2 - 5x - 8
that is, solve
x^2 - 5x - 8 = x - 1
x^2 - 6x - 7 = 0
(x-7)(x+1) = 0
x = -7 or x = -1
sub that back into y = x-1 to get
y = 6 or y = -2 respectively
so we have A(7,6) and B(-1,-2)
since these two points also lie on ty = p + qx - 2x^2 let's sub them in
6 = p + 6q - 98 and
-2 = p -q - 8
re-arrange these two equations and solve.
They are easy to solve as it by subtracting them, since the p's are the same.
Well, well, well! Looks like we have a straight line and a curve getting cozy on a graph. Let's see if we can help them find some common ground, shall we?
Now, since the straight line y = x - 1 and the curve y = x^2 - 5x - 8 both meet at points A and B, we can set their equations equal to each other to find those points. Here's what it looks like:
x - 1 = x^2 - 5x - 8
Now, we just need to solve this equation for x. Let's put on our math hats and get to work!
x^2 - 6x + 7 = 0
Factoring this, we get:
(x - 1)(x - 7) = 0
So, we have two possible values for x: x = 1 and x = 7. These are the x-coordinates of points A and B.
Now, we can substitute these x-values back into either the straight line or the curve equation to find the corresponding y-values. Let's use the straight line equation, y = x - 1, because it's easier to work with:
For x = 1:
y = 1 - 1 = 0
So, point A is (1, 0).
For x = 7:
y = 7 - 1 = 6
So, point B is (7, 6).
Now that we know points A and B, we can use them to find the values of p and q in the curve equation, y = p + qx - 2x^2. Let's do it step by step:
For point A (1, 0):
0 = p + q(1) - 2(1)^2
0 = p + q - 2
For point B (7, 6):
6 = p + q(7) - 2(7)^2
6 = p + 7q - 98
Now, we have a system of equations. Cue the dramatic music!
From the first equation, we have p = 2 - q.
Substituting this into the second equation:
6 = (2 - q) + 7q - 98
Simplifying:
6 = 9q - 96
9q = 102
q = 11.3333 (approximately)
Plugging this value of q back into p = 2 - q, we get:
p = 2 - 11.3333
p = -9.3333 (approximately)
So, the values of p and q are approximately -9.3333 and 11.3333, respectively.
And there you have it! The straight line and the curve have found their common ground. Now, go forth and spread the joy of math with these newfound values of p and q. Good luck!
To find the values of p and q, we need to equate the expressions for y from the straight line and the curve at points A and B.
First, let's find the x-coordinate of points A and B by setting the expressions for y equal to each other:
x - 1 = x^2 - 5x - 8
Rearranging this equation into quadratic form:
x^2 - 6x - 7 = 0
Factoring this quadratic equation:
(x - 7)(x + 1) = 0
Setting each factor equal to zero gives us two possible x-values:
x - 7 = 0 -> x = 7 (Point A)
x + 1 = 0 -> x = -1 (Point B)
Next, substitute these x-values into either the equation for the straight line or the curve to find the corresponding y-values.
For Point A (x = 7):
y = 7 - 1 = 6
For Point B (x = -1):
y = -1 - 1 = -2
Now, let's substitute the x and y-values for Point A into the equation for the curve to solve for p and q:
6 = p + q(7) - 2(7^2)
Simplifying:
6 = p + 7q - 98
Rearranging:
p + 7q = 104 (Equation 1)
Similarly, substitute the x and y-values for Point B into the equation for the curve:
-2 = p + q(-1) - 2(-1^2)
Simplifying:
-2 = p - q - 2
Rearranging:
p - q = 0 (Equation 2)
Now, solve these two equations simultaneously to find the values of p and q.
From Equation 2, we have p = q.
Substituting this into Equation 1:
p + 7p = 104
8p = 104
p = 13
Therefore, the values of p and q are p = 13 and q = 13.
To find the values of p and q, we need to substitute the coordinates of points A and B into the equation of the curve y = p + qx - 2x^2.
Let's start by finding the coordinates of points A and B where the straight line intersects the curve.
Given the straight line equation y = x - 1, we will equate it with the curve equation y = x^2 - 5x - 8.
Setting the two equations equal to each other, we have:
x - 1 = x^2 - 5x - 8
Rearranging and simplifying the equation:
x^2 - 6x - 7 = 0
Now, we can solve this quadratic equation. By factoring or using the quadratic formula, we find that the solutions are x = -1 and x = 7.
Substituting these x-values back into the straight line equation, we can find the corresponding y-values:
For x = -1:
y = -1 - 1 = -2
For x = 7:
y = 7 - 1 = 6
Therefore, the coordinates of point A are (-1, -2) and point B are (7, 6).
Now let's substitute these coordinates into the equation of the curve y = p + qx - 2x^2 to find the values of p and q.
For point A (-1, -2):
-2 = p + q(-1) - 2(-1)^2
-2 = p - q - 2
For point B (7, 6):
6 = p + q(7) - 2(7)^2
6 = p + 7q - 98
Now we have a system of two equations with two unknowns (p and q). Solving this system will give us the values of p and q.
From the equation for point A:
p - q - 2 = -2 (equation 1)
From the equation for point B:
p + 7q - 98 = 6 (equation 2)
We can solve this system of equations by substitution or elimination. Let's use elimination:
Multiply equation 1 by 7 to match the coefficient of q with equation 2:
7p - 7q - 14 = -14 (equation 3)
Add equation 3 to equation 2, eliminating q:
8p - 112 = -8
8p = 104
p = 13
Substitute the value of p back into equation 1 to find q:
13 - q - 2 = -2
-q = -13
q = 13
Therefore, the values of p and q are p = 13 and q = 13.
Hope this helps!