Question

How do you find an equation of the tangent line to the parabola `y=x^2-2x+7` at the point (3,10)?

Answer 1

`y=4x-2`

Explanation:

The gradient of the tangent to a curve at any particular point is give by the derivative of the curve at that point. The normal is perpendicular to the tangent, so the product of their gradients is `-1`

so If `y=x^2-2x+7` then differentiating wrt `x` gives us:

`dy/dx = 2x-2`

When `x=3 => y=9-6+7=10` (so `(3,10)` lies on the curve)
and `dy/dx=6-2=4`

So the tangent we seek passes through `(3,10)` ad has gradient `4` so using `y-y_1=m(x-x_1)` the equation we seek is;

`y-10=4(x-3)`
`:. y-10=4x-12`
`:. y=4x-2`

We can confirm this graphically: