[tex] f(x)=x^2 [/tex]
To find the gradient of tangent to a curve at a point we take the first derivative and then substitute the coordinates of the point in.
[tex] f'(x) = 2x [/tex]
[tex] f'(3) = 6 [/tex]
So this tells us the tangent has gradient 6. Now we know the function so we can work out the coordinates when x=3 since f(3) = 3²=9.
The tangent is a straight line with gradient 6 passing through (3,9) so
[tex] y-y_0=m(x-x_0) [/tex]
[tex] y-9=6(x-3) [/tex]
[tex] y=6x-9 [/tex]
review: power rule
[tex] \frac{d}{dx} x^n=nx^{n-1}[/tex]
use point slope form
[tex]y-y_1=m(x-x_1)[/tex]
where the slope is m and a point on the line is [tex](x_1, y_1)[/tex]
we can find the point by subsituting 3 for x
[tex]f(x)=x^2[/tex]
[tex]f(3)=(3)^2[/tex]
[tex]f(3)=9[/tex]
so the point is (3,9), ([tex]x_1=3[/tex] and [tex]y_1=9[/tex])
now we need the slope
do a bit of calculus
take the derivitive of f(x) to get f'(x) which is the slope of f(x) for any value of x
[tex]f(x)=x^2[/tex]
take derivitive, remember the power rule
[tex]\frac{d}{dx} x^2=2x^{2-1}=2x^1=2x[/tex]
therefore
f'(x)=2x
find slope at x=3
f'(x)=2x
f'(3)=2(3)
f'(3)=6
so the slope is 6 at that point
so we now know
m=6 and [tex](x_1,y_1)=(3,9)[/tex]
we can now write that the equation is
[tex]y-9=6(x-3)[/tex]
if you want it in other forms (I won't show work because if you are asking about derivitives, you should have your algebra down pat)
6x-y=9
y=6x-9
the equation is the ones below (same equation, different forms)
y-9=6(x-3)
6x-y=9
y=6x-9
