The function $y= 2x^2+2x+\frac{3}{2}$ is plotted below:


A non-horizontal straight line goes through point (2,1) and touches the curve. What is the slope of this tangent line?


The derivative of the quadratic is: $4x + 2$, we know that whatever tangent we have, it is going to have a slope of $(4x + 2)$ and it is going to go through the point (2,1): so $1 = 2(4x + 2) + c$.

Rearrange for c: $1 - 2(4x + 2) = c$, so the formula for the tangent is:

$y = x(4x + 2) + 1 - 2(4x + 2)$

This function, by the definition of the tangent must intersect our quadratic at point x. The the heights of the two functions must be equal at x, or:

$x(4x + 2) + 1 - 2(4x + 2) = 2x^2 + 2x + 3/2$

Note: the above quadratic has two solution:

$x = -1/2$ or $9/2$

but the -1/2 point will give you a horizontal tangent (which as you stated, you don't want!). To check this simply plug it in to the derivative: 4(-1/2) + 2 = 0 (hence the slope is zero, or i.e. the line is horizontal).

I trust you can plug in 9/2 to the correct equation in order to find the equation of the line ;)!

  • 1
    $\begingroup$ Thanks for the clear answer! $\endgroup$ – Bart Feb 16 '14 at 14:29

suppose that the line equation is $y-1=m(x-2)\implies y=mx-2m+1$. Suppose that the line touches the curve at $(a,b)$. Then $$m=y'(a)=4a+2 \tag{1}$$ $$b=2a^2+2a+3/2 \tag{2}$$ $$b=ma-2a+1 \tag{3}$$ Finally you solve equation 3 for $a$ using equations $1,2$


Suppose the line touches the curve at $(a,b)$. Then we have, from the equation of the curve:

$$b = 2a^2+2a+\frac32$$

The slope of the line is the derivative at $x=a$, which is $4a+2$. So the line is of the form

$$y-1 = (4a+2)(x-2)$$

This line goes through $(a,b)$, so we know that

$$b-1 = (4a+2)(a-2)$$


$$b = 1 + (4a+2)(a-2)$$

Now we have two equations for $b$. Setting them equal to each other gives

$$2a^2+2a+\frac32 = 1 + (4a+2)(a-2)$$

which is just a quadratic equation in $a$.


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