Question

The gradient of the curve y=2x^2-x+6 at the point (2,16)

Answer 1

The gradient of the curve y=2x^2-x+6 at the point (2,16)

At x = 2, y = 2(2)^2 - 2 + 6 = 2*4 + 4 = * + 4 = 12

The point (2,12) is a point on the curve y = 2x^2 - x + 6. The point (2,16) is not a point on this curve.

There is an error in the presentatiopn of the question (a typo)

We can show that the point (2.5,16) is a point on the curve y = 2x^2 - x + 6. (the origlnal curve)

And, the point (2,16) is a point on the curve y = 2x^2 + x + 6.  (note change of sign for the x-component)

So, either the coordinates are wrong, or the curve is wrong.

Getting the sign wrong is a frequent error, so I'm going to assume that's what's happened here.

Let the question be: The gradient of the curve y = 2x^2 + x + 6 at the point (2,16)

The point (2, 16) is a pont on the given curve.

The gradient of a curve, y = f(x), is given by dy/dx.

For the given curve, dy/dx = 4x + 1

At x = 2, dy/dx = 4*2 + 1 = 8 + 1 = 9

Answer: At the point (2, 16) the slope of the curve is 9

Answer 2

The given point lies inside the parabola (the given curve) so a line passing through the given point cannot be a tangent to the parabola. It’s possible that the y coord was calculated wrongly, since y=12 when x=2. (The given point does lie on the curve y=2x²+x+6.) The gradient is given by differentiating: y'=4x-1 which has the value 7 when x=2. If y'=4x+1 (differentiating y=2x²+x+6), the gradient is 9 when x=2.