Find the equation of the tangent at the point (2, 0) to the curve \(y = x^2 - 2x\)

Assessment: JAMB UTME - Mathematics - 2018 Subject: General Mathematics

Question 1 Report

y = 2x - 4

y = 2x + 4

y = 2x - 2

y = 2x + 2

Answer Details

The gradient to the curve is found by differentiating the curve equation with respect to x
So

\frac{d y}{d x}

2x - 2
The gradient of the curve is the same with that of the tangent.
At point (2, 0)

\frac{d y}{d x}

= 2(2) - 2
= 4 – 2 = 2
The equation of the tangent is given by (y - y1)

\frac{d y}{d x}

(x – x1)
At point (x1, y1) = (2, 0)
y - 0 = 2(x - 2)
y = 2x - 4

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