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Find the tangent to y = x^2 - 4x + 9 at the point (3,15)

First find dy/dx: dy/dx = 4x - 4 And thus at (3,15): dy/dx = 12 - 4 = 8 = m (as m is the gradient of a curve) So using y - y 1 = m(x - x 1 ) where (x 1 ,y 1 ) = (3,15): y - 15 = 8(x - 3) y = 8x- 9
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Answered by Scott H. • Maths tutor
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Find the derivative (dy/dx) of the curve equation x^2 -y^2 +y = 1.

Most of the differentiation problems require us to apply one of the well known rules, be it product rule, quotient rule or chain rule. But those problems have one thing in common: explicite formula for y, be...
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Answered by Adam G. • Maths tutor
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How do I solve a quadratic equation?

All quadratic equations can be written in the form ax 2 + bx + c = 0, where a, b and c are constants. Firstly, check whether you can easily factorise the equation into the form (x + p)(x + q) = 0, where pq =...
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Answered by Matthew S. • Maths tutor
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By completing the square, find any turning points and intersects with the x and y axes of the following curve. f(x) = 2x^2 - 12x +7

f(x) = 2x 2 - 12x +7 First take out a factor of 2 so that we have the coefficient of x 2 as 1. f(x) = 2[x 2 - 6x +7/2] Next, complete the sqaure on the part in square brackets. f(x) = 2[(x - 3) 2 - 9 + 7/2] ...
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Answered by Freddie I. • Maths tutor
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The equation of a line is y=3x – x^3 a) Find the coordinates of the stationary points in this curve, stating whether they are maximum or minimum points b) Find the gradient of a tangent to that curve at the point (2,4)

a) A stationary point is any point on the curve that is flat, still, not increasing or decreasing. Another way to think of this is the gradient at a stationary point = 0 Firstly make an equation for your gra...
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Answered by Lauren C. • Maths tutor
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