Question

A quadratic function is defined by
f(x)= 3x^2+4x-2. A linear function is defined by g(x)= mx-5. what values(s) of the slope of the line would make it a tangent to the parabola.

Answer 1

Slope of any function is first derivate of that function.

Quadratic function in your case have first dervate:

3*2*x+4=6*x+4

Linear function in your case have first dervate:

m

Line will be tangent of quadratic parabola when first derivates is egual.

m=6*x+4
m*x-5=(6*x+4)*x-5=6*x^2+4*x-5

3*x^2+4*x-2=6*x^2+4*x-5
3*x^2+4*x-6*x^2-4*x=-5+2
-3*x^2=-3 Multiply with (-1)
3*x^2=3 Divided with 3
x^2=1
x=+OR-(1)
x1=-1 , x2=1

Slope for x1=6*x1+4=6*(-1)+4=-6+4=-2

Slope for x2=6*1+4=6+4=10

S1=-2 S2=10

Answer 2

How can i get 6*x+4?