1. ## Mathematics Ib 3283 quadratic question

Find the values of m for which the lines y=mx-2 are tangents to the curve with equation y=x^2-4x+2

2. ## Re: Mathematics Ib 3283 quadratic question

the y-values of a tangent line and curve are equal at the point of tangency, plus their respective slopes are also equal ...

3. ## Re: Mathematics Ib 3283 quadratic question

Ok so what does that mean, how would I find m?

4. ## Re: Mathematics Ib 3283 quadratic question

Originally Posted by alexj987
Ok so what does that mean, how would I find m?
exactly what I stated:

y-values are equal ...

$mx-2 = x^2-4x+2$

and slopes are equal ...

$\frac{d}{dx}(mx-2) = \frac{d}{dx}(x^2-4x+2)$

5. ## Re: Mathematics Ib 3283 quadratic question

Yes, Ive made the two equations equal each other earlier but still do not fully understand, sorry for being dull ...

6. ## Re: Mathematics Ib 3283 quadratic question

have you not solved a system of equations before?

$mx-2 = x^2-4x+2$

$m = 2x-4$

substitute $(2x-4)$ for $m$ in the first equation, solve for $x$, then determine the possible value(s) for $m$

7. ## Re: Mathematics Ib 3283 quadratic question

Did that, factored and got an answer of x=1, then plugged that back into x^2-4x+2 which gives -1

8. ## Re: Mathematics Ib 3283 quadratic question

your value of x is incorrect ... check your algebra. Also, you get two possible values for x.

9. ## Re: Mathematics Ib 3283 quadratic question

Yes, Im sorry, ok using the quadratic formula, x=1+/-√2

10. ## Re: Mathematics Ib 3283 quadratic question

$(2x-4)x - 2 = x^2 - 4x + 2$

$2x^2 - 4x - 2 = x^2 - 4x + 2$

$x^2 - 4 = 0$

try again ...