Can someone help me with this problem please? (Thanks in advance):
The graph of $y=3x$ is tangent to $f(x)=2x^2-ax+b$ at $x=2$. Find the values of $a$ and $b$.
If $y=3x$ is tangent to $f(x)$ at $x=2$, then $f(x)$ has slope $3$ at $x=2$. To find the slope, we differentiate $f$ and evaluate at 2 to get $3=f'(2)=8-a$, so $a=5$. To find $b$, we note that if $f$ is tangent to $y=3x$ at $x=2$, then it must pass through $(2,y(2))=(2,6)$. Setting $f(2)=2\cdot 2^2 -5 \cdot 2 + b = 6$, we get $b=8$.
A few hints:
1) Given a curve $f(x)$, how do you compute the slope at a point $P$ on the curve? If $y = 3x$ is a tangent at P, what is the slope at point P from this information?
2) If the line $y = 3x$ and the curve $y = 2x^2 - ax + b$ meet, they have to meet at exactly one point. So what can you say about the roots of the quadratic equation $3x = 2x^2 - ax + b$?