To find the equation of the tangent, we need a gradient and the coordinates of a point.

To find the y-coordinate, we substitute the value of x into the original equation.

To find the gradient, we perform dy/dx on the original equation before substituting the value of x.

A normal to a curve is a line which is perpendicular to the tangent to the curve at the same tangential point. The product of gradients of two perpendicular lines is -1, so you can use the previous result to find the gradient of the normal.

With these, you are ready to form the equations of the tangent and normal.

I have successfully found the equation of the tangent but I can’t get the correct answer for the equation of normal. my answer to equation of tangent is y=47x-97 but I’m not sure why my equation to normal is wrong

Oh sorry, I didn’t read the question carefully. They wishes for a normal at a different point. Let me take a look…

x = -2, y = 5

Coordinates of point: (-2, 5)

dy/dx
= 8x + 7
= 8 (-2) + 7
= -9

Note that dy/dx always represents the gradient of the tangent, not the normal.

Gradient of tangent at x = -2 is -9.
Gradient of normal at x = 2 is 1/9.

For a line of gradient 1/9 passing through (-2, 5),

y = (1/9) x + c
5 = (1/9) (-2) + c
47/9 = c

So, the equation of the normal at x = -2 is

y = (1/9)x + 47/9
9y = x + 47