KCSE 2025 Maths P1 Q16 — Tangent to a Curve

KCSE 2025 Form 4 Calculus

Published

The Question

“The curve y = 3x^2 - 4x passes through the point P(1, -1). Determine the equation of the tangent to the curve at P, leaving the answer in the form ax + by = c, where a, b and c are scalars.”

1

Differentiate the curve

The gradient of a curve at any point is given by its derivative. Differentiate the equation of the curve term by term to get a formula for the gradient at any value of x.

y=3x2−4xy = 3x^{2} - 4x
dydx=6x−4\frac{dy}{dx} = 6x - 4
2

Find the gradient at P

Substitute the x-coordinate of P, which is 1, into the derivative. This gives the exact slope of the tangent line where it touches the curve at P.

m=6(1)−4=2m = 6(1) - 4 = 2
3

Form the tangent line through P

The tangent is a straight line passing through P with the gradient just found. Use the point-slope form of a line, substituting the coordinates of P and the gradient.

y−y1=m(x−x1)y - y_{1} = m(x - x_{1})
y−(−1)=2(x−1)y - (-1) = 2(x - 1)
y+1=2x−2y + 1 = 2x - 2
4

Rearrange to the required form

Simplify and move the terms so the equation matches the form ax + by = c that the question asks for.

y=2x−3y = 2x - 3
2x−y=32x - y = 3

Final Result

The equation of the tangent to the curve at P is 2x - y = 3.

Why this method works

A tangent touches a curve at a single point and has the same steepness as the curve there. Because the derivative measures how fast y changes with x, evaluating dy/dx at that point gives the exact gradient of the tangent. Once you know a gradient and one point the line passes through, the point-slope form fixes the line uniquely, so rearranging it simply presents the same line in the format the examiner requested.

✓

Substitute P(1, -1) into 2x - y = 3: 2(1) - (-1) = 2 + 1 = 3, which is true, confirming the line passes through P.