The equation of the tangent to the curve at the point P(2, -10) is:
y = 8x - 26
We need to find the coordinates of point Q where the slope of the tangent to the curve f(x) must also be 8.
Now we have a quadratic:
which simplifies to:
which factorizes to:(x + 4)(x - 2) = 0
Therefore x = -4, 2.
f(-4) = 48
Therefore the coordinates of Q are (-4, 48).
y = 8x - 26
We need to find the coordinates of point Q where the slope of the tangent to the curve f(x) must also be 8.
Now we have a quadratic:
which simplifies to:
which factorizes to:(x + 4)(x - 2) = 0
Therefore x = -4, 2.
f(-4) = 48
Therefore the coordinates of Q are (-4, 48).