# Thread: Equation of line that is tangent to graph of f and parallel

1. ## Equation of line that is tangent to graph of f and parallel

I need to find an equation of the line that is tangent to the graph and parallel to the given line:

function:
$f(x)=x^2-x$

line:
$x+2y-6=0$

How can I do this problem? (step by step always works best for me )

2. Good day AMaccy. Step by step it is then

Step 1. Rearrange the line equation into the form y=mx+c
$y=\frac{-x}{2}+3$

Step 2. Remember that your teacher said,for our line to be parallel to the given line, it must have the same gradient.

Step 3. Gain insight/remember for a line to be tangent to a curve, it must possess the same gradient as the curve at one particular point.

Step 4. Knowing these 2 facts, we state that the gradient of our wanted line is $m=\frac{-1}{2}
$

Step 5. Differentiate the curve to discover where it's gradient is $\frac{-1}{2}$
$\frac{dy}{dx}=2x-1$, Let $\frac{dy}{dx}=2x-1=\frac{-1}{2}$
$x=\frac{1}{4}
$

Step 6. Obtain the y-coordinate at x=
$f(\frac{1}{4})=(\frac{1}{4})^2-\frac{1}{4}$
$=\frac{-3}{16}$

Step 7. We now have the point at which the curve and line touch and our line's gradient, calculate our line's y-intercept.
$y=mx+c$
$\frac{-3}{16}=\frac{-1}{2}(\frac{1}{4})+c$
$c=\frac{-1}{16}$

So the equation of the line is $y=\frac{-1}{2}x-\frac{-1}{16}$

3. Originally Posted by AMaccy
I need to find an equation of the line that is tangent to the graph and parallel to the given line:

function:
$f(x)=x^2-x$

line:
$x+2y-6=0$

How can I do this problem? (step by step always works best for me )

1. Re-write the equation of the line:

$y=-\dfrac12 x +3$ . Thus the slope is $m = -\dfrac12$

2. The slope of the tangent equals first derivation of f at the tangent point. Therefore you need the first derivation:

$f'(x)= 2x-1$

Solve for x: m = f'(x):
$-\dfrac12=2x-1~\implies~x=\dfrac14$

The tangent point has the coordinates $T\left(\dfrac14\ ,\ -\dfrac3{16}\right)$ because T is a point of the graph of f.

3. Use the point-slope-formula to get the equation of the tangent:

$y-\left(-\dfrac3{16}\right)=-\dfrac12 \left(x-\dfrac14\right)$ Solve for y:

$y=-\dfrac12 x -\dfrac1{16}$