Work out the equation of the tangent to the curve y=x^2+5x-8 at the point (2,6)

Firstly, we know that dy/dx is the gradient of the tangent to a curve. If we know the gradient of the tangent, and the point on the curve where the tangent touches it, we can work out the equation of the tangent.Now, we will find dy/dx. If y=xⁿ, then we know that dy/dx=nx^n-1So if y= x²+5x-8, then dy/dx=2x+5.At the point (2,6), where x=2, the gradient of the tangent is dy/dx= 2(2)+5=9. The general equation of a straight line is y=mx+c, where m is the gradient.So, to find the equation of the tangent, we will sub in m=9. So y=9x+c.At the point (2,6), x=2 and y=6, which implies 6=9(2) +c, so c=-12.Therefore, the equation of the tangent is y=9x-12.