1. ## The tagents line

Hi all, I have a problem here and I don't know how to find a tangent line. Some hints may help me. Thank you so much.
Let f(x) = $x^2-x+3$
Find the straight lines that are tangent to the graph of f and going through the point (3/4,11/4)

2. ## Re: The tagents line

differentiate f(x) you will get d(f(x))=2x-1

now substitute 3/4,11/4 in it as

11/4=2(3/4)-1+c

find c
thus you will get a tangent line

y=2x-1+c
where c is the value which was found above

3. ## Re: The tagents line

I have the solution but I don't understand it. Can you explain it ?

f'(x)=2x-1=(y-11/4)/(x-3/4)
Therefore (2x -1)(x-3/4)=( $x^2+x-3$)-11/4 or $x^2- 3/2 x + 1/2$=0
For x=1/2 the slope 2x-1=0 => the tangent line is y=11/4
For x=1 the slope 2x-1=1 => the tangent line is y=x+2

4. ## Re: The tagents line

the last two points are obvious as (2x-1)=(y-11/4)/(x-3/4)
puting x=1/2 in above will give y=11/4
putting x=1 in above will give y=x+2
and we have cross multiplied (x-3/4) with (2x-1) from above eqn
so we get

2x^(2)-(3/2)x-x+3/4=y-11/4

now substituting y=x^(2)-x+3

we get 2x^(2)-(3/2)x-x+3/4=x^(2)-x+3-11/4

now solve and u will get

x^(2)-(3/2)x+1/2

now as slope of the function has to be found out on 0 so equate the above term with 0

use breaking of middle terms so u split 3/2 into 1 and 1/2

so u get

(x-1/2)(x-1)=0

so x=1/2 or x=1

hence u get the above result

5. ## Re: The tagents line

Originally Posted by math88
I have the solution but I don't understand it. Can you explain it ?

f'(x)=2x-1=(y-11/4)/(x-3/4)
Therefore (2x -1)(x-3/4)=( $x^2+x-3$)-11/4 or $x^2- 3/2 x + 1/2$=0
For x=1/2 the slope 2x-1=0 => the tangent line is y=11/4
For x=1 the slope 2x-1=1 => the tangent line is y=x+2
$f(x) = x^2 - x + 3 \implies f'(x) = 2x - 1.$

Let's say the tangent to f(x) when x = a (a is our unknown) runs through (3/4, 11/4). So the tangent's slope is 2a - 1.

$\dfrac{\frac{11}{4} - (a^2 - a + 3)}{\frac{3}{4} - a} = 2a - 1 \implies\dfrac{4\left(\frac{11}{4} - a^2 + a - 3\right)}{4\left(\frac{3}{4} - a\right)} = 2a - 1 \implies \dfrac{11 - 4a^2 + 4a - 12}{3 - 4a} = 2a - 1 \implies$

$- 1 - 4a^2 + 4a = 6a - 3 - 8a^2 + 4a \implies 4a^2 - 6a + 2 = 0 \implies a = \dfrac{- (-6) \pm \sqrt{(-6)^2 - 4 * 4 * 2}}{2 * 4} = \dfrac{6 \pm \sqrt{36 - 32}}{8} = 1\ or\ \dfrac{1}{2}.$

At this point we have the x values for the points of tangency.

Case I: $a = 1.$

Slope = $2a - 1 = 2 * 1 - 1 = 2 - 1 = 1.$

y value = $1^2 - 1 + 3 = 3$.

Equation is $\dfrac{y - 3}{x - 1} = 1 \implies y - 3 = x - 1 \implies y = x + 2.$

Check $y = x + 2 \implies y = \dfrac{3}{4} + 2 = \dfrac{11}{4}\ when\ x = \dfrac{3}{4}.$

Case II: $a = \dfrac{1}{2}.$

Slope = $2a - 1 = 2\left(\dfrac{1}{2}\right) - 1 = 1 - 1 = 0.$

y value = $\left(\dfrac{1}{2}\right)^2 - \dfrac{1}{2} + 3 = \dfrac{1}{4} - \dfrac{1}{2} + 3 = 3 - \dfrac{1}{4} = \dfrac{11}{4}.$

Equation is $\dfrac{y - \frac{11}{4}}{x - \frac{1}{2}} = 0 \implies y - \dfrac{11}{4} = 0 \implies y = \dfrac{11}{4}.$

Check $y = \dfrac{11}{4} \implies y = \dfrac{11}{4}\ when\ x = \dfrac{3}{4}.$

6. ## Re: The tagents line

Why can't I solve the problem like this?
The slope m=f’(x) =2x-1 =2. ¾ -1=1/2
The tangent line is y-11/4=1/2(x-3/4 => y=1/2 x + 19/8

7. ## Re: The tagents line

Originally Posted by math88
Why can't I solve the problem like this?
The slope m=f’(x) =2x-1 =2. ¾ -1=1/2
The tangent line is y-11/4=1/2(x-3/4 => y=1/2 x + 19/8
That would work IF the given point, (3/4, 11/4), were a point on the curve. But it isn't as you can see by setting x= 3/4 in $x^2- x+ 3$. The result is 45/16, not 11/4.

You are asked to find the equation of a line that passes through a given point that is NOT on the curve but the line is tangent to the curve at some other point.