Question 1071734
<pre>
The other tutor assumed you are taking calculus, not 
just pre-calculus or college algebra.  I will use
only the mothods taught in algebra and pre-calc in 
case you aren't taking calculus.  It's easier with
calculus, but can be done with algebra only.

The point-slope form of a line is

{{{y-y[1]=m(x-x[1])}}}

If the line goes through (2,5) we can substitute 
(x<sub>1</sub>,y<sub>1</sub>)=(2,5) and we have:

{{{y-5=m(x-2)}}}

{{{y-5=mx-2m}}}

{{{y=mx-2m+5}}}

We substitute mx-2m+5 for y in

{{{mx-2m+5 = x^2-3x+7}}}

{{{mx-2m = x^2-3x+2}}}

{{{0=x^2-mx-3x+2m+2}}}

{{{x^2-(m+3)x+2(m+1)=0}}}

In order for the line to be tangent,
the discriminant must = 0

{{{b^2-4*a*c=0}}}
{{{(-(m+3))^2-4*(1)*2(m+1)=0}}}
{{{(m+3)^2-8(m+1)=0}}}
{{{m^2+6m+9-8m-8=0}}}
{{{m^2-2m+1=0}}}
{{{(m-1)(m-1)=0}}}
{{{(m-1)^2=0}}}
{{{m-1=0}}}
{{{m=1}}}

So the line's equation

{{{y=mx-2m+5}}}

is now

{{{y=1x-2*1+5}}}

{{{y=x-2+5}}}

{{{y=x+3}}}

{{{drawing(192,400,-1.5,4.5,-2.5,10,

graph(192,400,-1.5,4.5,-2.5,10,x^2-3x+7),
line(-6,-3,6,9), locate(2,5,"(2,5)") )}}}

Edwin</pre>