Find the equation of the straight line tangent to the curve y=2x^3+3x^2-4x+7, at the point x=-2.
We are looking for a straight line, so it needs the form y=mx+c. To find our gradient, m, we need the gradient of the curve at the point x=-2, so differentiate the equation: dy/dx=6x2+6x-4, and solve at x=-2, ie m=64-62-4=8.To find c, calculate the y coordinate at x=-2 using the equation of the curve: y=2*(-8)+34-4(-2)+7=11. Using the values we have for y, m and x, we can calculate what value c should be: y=mx+c, so c=y-mx=11-8*(-2)=27.Thus the equation for the tangent line at x=-2 is y=8x+27.
JB
Related Maths A Level answers
All answers ▸The curve C has equation y = 3x^4 – 8x^3 – 3. Find dy/dx.
Consider the unit hyperbola, whose equation is given by x^2 - y^2 = 1. We denote the origin, (0, 0) by O. Choose any point P on the curve, and label its reflection in the x axis P'. Show that the line OP and the tangent line to P' meet at a right angle.
Solve the following equation: x^3 + 8x^2 + 4x - 48=0
A curve has an equation: (2x^2)*y +2x + 4y – cos(pi*y) = 17. Find dy/dx