I'm in an American high school but my teacher gave me some IBDP questions and I'm struggling to solve one of them:
Find the exact values of $m$ for which the line $y= mx+3$ is tangent to the curve with equation $y=3x^2 -x +5$
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What does it mean for $y=mx+3$ to be tangent to $y=3x^2-x+5$? It means the quadratic equation $3x^2-x+5-(mx+3)=0$ has exactly one solution, or equivalently, the discriminant is zero, so...
Edit: as per @Glen O's comment, I find it necessary to clarify a little bit how my method works.
Let $f(x)$ denote the quadratic function. Then $f'(x)$ (the slope) is a nonzero linear function, so is strictly monotone, or more generally, is injective on $\Bbb R$. Now given a line $l$ whose slope is $k$, suppose it intersects the quadratic function curve at two distinct points $a,b$ with $a<b$, then by Lagrange MVT, there exists $c\in (a,b)$ such that $f'(c)=k$. But $f'(x)$ is injective, so neither $f'(a)$ nor $f'(b)$ equals $k$, so $l$ must not be tangent to $y=f(x)$.
We need to find $x$ values for which the slopes are the same, and the $y$ values are the same. That's what tangent means. Let $f(x) = mx + 3$ and let $g(x) = 3x^2 - x + 5$.
First take the derivative of $g(x)$, which is $$ g'(x) = 6x - 1. $$ This will give you the slope at any given point.
We want $m$ to be equal to $g'(x)$, so that they have the same slope. We also want $f(x)$ to equal $g(x)$. So we have the two equations: $$ m = 6x - 1 $$ and $$ mx + 3 = 3x^2 - x + 5. $$ We can substitute $6x - 1$ for $m$ in the second equation $$ (6x - 1)x + 3 = 3x^2 - x + 5 $$ and solve for $x$.
OK so I started off by setting the equations equal to each other, which gave me:
$mx + 3 = 3x^2 -x +5$
I put the 3 on the other side:
$mx = 3x^2 -x +2$
Then I set it equal to zero:
$3x^2 + x(1-m) +2 = 0$
And I'm stuck now. I tried using the discriminant or the quadratic formula but I got really complicated answers and they didn't seem right. I tried solving it for the past 30 minutes and I came up with no answer.
Any input would be appreciated.