Question: Find the co-ordinates and the gradient of two intersecting points on different equations.?
I would of showed my attempts to work this out, but I'm completely stumped on this question and have no idea how to work it out.
Could someone show me how to work this exercise out?
The straight line y=8x+16 cuts the curve y=8x^2 at two points. Find the co-ordinates of each point and the gradient of the curve at each one.
Relation Questions:
Answer:
y = 8x + 16 = 8x^2 (at the intersections)
8x^2 - 8x - 16 = 0
x^2 - x - 2 = 0
(x - 2)(x + 1) = 0
so either x = 2 or x = -1
when x = 2, then y = 32 ==> (2 , 32)
when x = -1, then y = 8 ==> (-1 , 8)
y = 8x^2 ==> y' = 16x
at x = 2, then y' = 32 = slope of gradient at (2 , 32)
at x = -1, then y' = -16 = slope of gradient at (-1 , 8)
Hi,
At their intersections y = 8x + 16 equals y = 8x²
8x² = 8x + 16
8x² - 8x - 16 = 0
8(x² - x - 2) = 0
8(x - 2)(x + 1) = 0
x = 2 and x = -1 are the x values of the 2 intersection points.
The 2 intersection points are (2,32) and (-1, 8). <==ANSWER
If you are asking for the gradient of y = 8x² at those 2 points, then the first derivative is f'(x) = 16x.
The gradient of y = 8x² at (2,32) is 16(2) or 32. <==ANSWER
The gradient of y = 8x² at (-1,8) is 16(-1) = -16. <==ANSWER
I hope that helps!! :-)
(1) : y = 8x + 16 → this is the line
(2) : y = 8x² → this is the curve
line = curve → point of intersection
y = y
8x² = 8x + 16 → you can simplify by 8
x² = x + 2
x² - x - 2 = 0
Polynomial like: ax² + bx + c, where:
a = 1
b = - 1
c = - 2
Δ = b² - 4ac (discriminant)
Δ = (- 1)² - 4(1 * - 2) = 1 + 8 = 9 = 3²
x1 = (- b - √Δ) / 2a = (1 - 3) / (2 * 1) = - 1
x2 = (- b + √Δ) / 2a = (1 + 3) / (2 * 1) = 2
Recall: (2) : y = 8x²
When x = x1 = - 1 → y = 8 → the coordinates are: (- 1 ; 8)
When x = x2 = 2 → y = 32 → the coordinates are: (2 ; 32)