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 ▸Solve the simultaneous equations: y-2x-4 = 0 (1) , 4x^2 +y^2 + 20x = 0 (2)
Given f(x): 2x^4 + ax^3 - 6x^2 + 10x - 84, and knowing 3 is a root of f(x), which is the value of a?
Mechanics (M1): Particle moving on a straight line with constant acceleration (Relationships of the 5 Key Formulae)
Find the gradient of the tangent to the curve with the equation y = (3x^4 - 18)/x at the point where x = 3