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:
$\displaystyle f(x)=x^2-x$

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

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
$\displaystyle y=\frac{-x}{2}+3$

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 $\displaystyle m=\frac{-1}{2}$
Step 5. Differentiate the curve to discover where it's gradient is $\displaystyle \frac{-1}{2}$
$\displaystyle \frac{dy}{dx}=2x-1$, Let $\displaystyle \frac{dy}{dx}=2x-1=\frac{-1}{2}$
$\displaystyle x=\frac{1}{4}$
Step 6. Obtain the y-coordinate at x=
$\displaystyle f(\frac{1}{4})=(\frac{1}{4})^2-\frac{1}{4}$
$\displaystyle =\frac{-3}{16}$

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.
$\displaystyle y=mx+c$
$\displaystyle \frac{-3}{16}=\frac{-1}{2}(\frac{1}{4})+c$
$\displaystyle c=\frac{-1}{16}$

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

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:
$\displaystyle f(x)=x^2-x$

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

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

1. Re-write the equation of the line:

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

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

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

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

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

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

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

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