Find constants so a quadratic has a given tangent at a particular point

by RoRi

September 17, 2015

Let $f$ be the quadratic function

$f(x) = x^2 + ax + b \qquad a,b \text{ constants in } \mathbb{R}.$

Find values for $a$ and $b$ such that the graph of $f$ has a tangent line given by $y = 2x$ at the point $(2,4)$ .

First, we compute the derivative

$f'(x) = 2x + a.$

So, if the line $y = 2x$ is tangent to $f$ at the point $(2,4)$ we must have

$f'(2) = 2 \quad \implies \quad 4+a = 2 \quad \implies \quad a = -2.$

Next, the point $(2,4)$ must be on the graph of $f$ , i.e., we must have $f(2) = 4$ . Thus,

$f(2) = 4 \quad \implies \quad 4 + (-2)2 + b = 4 \quad \implies \quad b = 4.$

So, the values are $a = -2$ and $b = 4$ .

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4 comments

August 18, 2019 at 6:26 am

varshini says:

why is f'(2)=2 ?

Reply
- September 16, 2020 at 3:16 am
  
  Mihajlo says:
  
  Because the tangent at point (2, 4) is 2 (i.e. the slope of line 2x is 2).
  
  Reply
May 17, 2016 at 11:07 pm

Anonymous says:

I think there might be an error in this answer, why do you have f'(2) = 2?
is asking for the the tangent to be equal to y = 2x. f'(x) = 2x + a, so at (2,4) a should = 0.
and therefore b = 0 as well…

Reply
- May 18, 2016 at 9:32 am
  
  RoRi says:
  
  No, the answer is correct. It is asking for the tangent line to be equal to $y = 2x$ at the point $(2,4)$ , not for the derivative to be $2x$ . (The derivative gives the slope of the tangent line, it is not a formula for the tangent line.) So, since $y = 2x$ is tangent to the curve at $(2,4)$ and this line has slope 2, we know the derivative must be equal to 2 at that point. Hence, $f'(2) = 2$ . Does that make sense?
  
  Reply

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