# 7. Let P And Q be points on the curve y=x^2+2x+5 at which x=2 and x=2+h respectively. Express the gradient of the line PQ in terms of h, and hence find the gradient of the tangent to the curve y=x^2+2x+5 at x=2. Let P And Q be points on the curve y=x^2+2x+5 at which x=2 and x=2+h respectively.

Step-by-step solution-:

Simplify the given expression for x=2:

y=x^2+2x+5

Simplify the given expression for x=2+h:

y=(2+h)^2+2(2+h)+5

Let be m a slope of line PQ:

The slope formula is:

m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}

Substitute the coordinate of points A(2,13) and B(2+h, h^2+6h+13) into the slope formula:

m=\frac{h^2+6h+13-13}{2+h-2}

After simplifying we get,

m=\frac{h^2+6h}{h}
m=\frac{\cancel{h}(h+6)}{\cancel{h}}

After simplifying we get,

m=h+6

We know that the first derivative of any curve at the point (x,y) is equal to the slope of the tangent:

It means,

m=\frac{dy}{dx}

Substitute the y=x^2+2x+5 into the above:

m=\frac{d(x^2+2x+5)}{dx}=2x+2

Substitute the y=2 into the above:

m=2\times2+2

After simplifying we get,

m=6

You may be like more post-: