# Math Help - [SOLVED] Is this question possible without having learnt derivatives?

1. ## [SOLVED] Is this question possible without having learnt derivatives?

"For what value of b will the line y=-2x+b be tangent to the parabola y=3x^2+4x-1?"

I can get the answer through some trial and error, but that's not really math. Looking it up online, all the methods use derivatives. In my class though, we haven't learnt derivatives.

Attempt:

1) y=-2x+b
2) y=3x^2+4x-1

Substitute 1 into 2.

-2x+b=3x^2+4x-1
0=3x^2+6x-1-b
now for the trial and error part:
b must be such that the value that b gives c, when divided by a, has a perfect square root, so that there is only one value for x (because it is a tangent)
ex: 0=3x^2+6x-1-(-4)
0=3(x^2+2x+1)
0=3(x+1)^2
now you have to make sure with a value of -4 for b, it actually intercepts the parabola, which is does.
but I could have also got -28 for b, which would work with the first part, but does not intercept the parabola, so a lot of trial and error with this method.

2. Originally Posted by yetuop
"For what value of b will the line y=-2x+b be tangent to the parabola y=3x^2+4x-1?"

I can get the answer through some trial and error, but that's not really math. Looking it up online, all the methods use derivatives. In my class though, we haven't learnt derivatives.

Attempt:

1) y=-2x+b
2) y=3x^2+4x-1

Substitute 1 into 2.

-2x+b=3x^2+4x-1
0=3x^2+6x-1-b
now for the trial and error part:
b must be such that the value that b gives c, when divided by a, has a perfect square root, so that there is only one value for x (because it is a tangent)
ex: 0=3x^2+6x-1-(-4)
0=3(x^2+2x+1)
0=3(x+1)^2
now you have to make sure with a value of -4 for b, it actually intercepts the parabola, which is does.
but I could have also got -28 for b, which would work with the first part, but does not intercept the parabola, so a lot of trial and error with this method.
The line is a tangent if and only if the quadratic

$3x^2+6x-(1+b)=0$

has exactly one real root, which is the case when the discriminant is zero:

$36+12(1+b)=0$

which is the case when $b=-4$.

CB