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 of the slope of the line would make it a tangent to the parabola?

Answer 1

After I equated the function, I got 3x^2 + 4x - mx + 3 = 0.
b^2 -4ac = 0
Is b = 4-m?

Answer 2

heelp D:

Answer 3

The answer is supposed to be m = 10, m = -2

Answer 4

3x^2 + 4x - 2= mx-5
3x^2 + (4-m)x + 3=0
a=3 b=4-m c=3

b^2 -4ac = 0

Answer 5

why is it (4-m)x?

Answer 6

that's the thing that's confusing me..

Answer 7

3x^2 + 4x - 2= mx-5
3x^2 + 4x - 2 -mx +5=0
3x^2 + 4x -mx + 3= 0
3x^2 + x(4-m) + 3=0

Answer 8

common factor
4x -mx
x(4-m)

Answer 9

ohhh you factored it. Thank you! :')

Answer 10

so I don't put b^2 - 4ac = x(4-m)^2 - 4(3)(3)?

Answer 11

without the x

Answer 12

why not?

Answer 13

(4-m)^2 - 4(3)(3)=0

Answer 14

you want the derivative of the quadratic to have the same slope as the linear equation and pass thru the point 0,5

Answer 15

what is the linear equation's slope?

Answer 16

y = f'(0)x - 5 is the equation of the line that you want to equate

f(x) = 3x^2 + 4x - 2
f'(x) = 6x +4

f'(0) = 0+4 = 4

y = 4x - 5 ; so the points on the parabola with a slope of 4

Answer 17

ideally there should be 2 values for the slope of this line

Answer 18

3x^2 + 4x - 2 = mx - 5

3x^2 + 4x - 2 = mx - 5
-mx+5 -mx+5
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3x^2+(4-m)x +3 = 0 looks to be the route sam was going

Answer 19

-(4-m) +- sqrt((4-m)^2 - 4(3)(3))
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2(3)


m-4 +- sqrt(16+m^2-8m - 36)
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6


m-4 +- sqrt(m^2-8m - 20)
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6