SOLUTION: Topic : Differentiation (Q) The line (l) is the tangent, at the point (2,3), to the curve with equation y = a + bx^2, where a and b are constants. The tangent (l) has gradient 8

Question 565992: Topic : Differentiation
(Q) The line (l) is the tangent, at the point (2,3), to the curve with equation y = a + bx^2, where a and b are constants. The tangent (l) has gradient 8. Find the values of a and b ?
Answer by htmentor(1343) (Show Source): You can put this solution on YOUR website!
(Q) The line (l) is the tangent, at the point (2,3), to the curve with equation y = a + bx^2, where a and b are constants. The tangent (l) has gradient 8. Find the values of a and b ?
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The slope (gradient) of the tangent line to the curve y = a + bx^2 is
dy/dx = 2bx = 2b(2) = 4b
Since the gradient is 8, we have the value for b:
4b = 8 -> b = 2
Now use the equation for the curve to solve for a:
(x,y) = (2,3)
3 = a + 2*2^2
3 = a + 8
This gives a = -5
So the curve is y = -5 + 2x^2
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