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February 1, 2015

February 1, 2015

Posted by **priya** on Wednesday, November 3, 2010 at 11:54pm.

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.

- math grade 11 -
**Bosnian**, Thursday, November 4, 2010 at 2:40amSlope 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

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