# Math Help - equation of tangent #1

1. ## equation of tangent #1

Find the equation of the tangents to the curve y=2x^2+3 that passes through the following:
a) point (2,3)

my work is below:
y=2x^2+3 PP(2,3)
y'=4x
at x=2
f'(2)=4(2)
=8

y=8x+b
3=8(2)+b
b=-13

y=8x-13

Can someone please verify or check to see if my answer or the way that i did it is correct? Thanks

2. Originally Posted by skeske1234
Find the equation of the tangents to the curve y=2x^2+3 that passes through the following:
a) point (2,3)

my work is below:
y=2x^2+3 PP(2,3)
y'=4x
at x=2
f'(2)=4(2)
=8

y=8x+b
3=8(2)+b
b=-13

y=8x-13

Can someone please verify or check to see if my answer or the way that i did it is correct? Thanks

the function dose not pass through that point

$f(x)=2x^2+3 \Rightarrow f(2)=2(2)^2+3 = 8+3=11$

the function pass through (2,11) not (2,3)

3. Originally Posted by Amer
the function dose not pass through that point

$f(x)=2x^2+3 \Rightarrow f(2)=2(2)^2+3 = 8+3=11$

the function pass through (2,11) not (2,3)
so what did i do wrong and how do i fix it?

4. Originally Posted by skeske1234
so what did i do wrong and how do i fix it?
You've done nothing wrong, you got a faulty question. Did you type it in properly?

5. Originally Posted by skeske1234
so what did i do wrong and how do i fix it?
the wrong is in the question not in your solution since the curve dose not pass through (2,3) so you can't find the tangent for the curve at this point but you can find the tangent at the point (2,11) since the curve pass through it

6. Originally Posted by e^(i*pi)
You've done nothing wrong, you got a faulty question. Did you type it in properly?
I think it means this:
Given the parabola y=2x^2 +3 find the equation of the tangents that passes through the point a) (2,3) and b) (2,-7)

7. Originally Posted by skeske1234
I think it means this:
Given the parabola y=2x^2 +3 find the equation of the tangents that passes through the point a) (2,3) and b) (2,-7)
Yeah, it's an impossible/wrong question.

As Amer shows in post 2 neither (2,3) nor (2,-7) lie on the line

8. Originally Posted by skeske1234
Find the equation of the tangents to the curve y=2x^2+3 that passes through the following:
a) point (2,3)
the question asked for the lines tangent to the curve that pass thru the point (2,3) which is not on the curve.

point on the curve ... $(x, 2x^2+3)$

slope between the point on the curve and the point (2,3) ...

$m = \frac{(2x^2+3) - 3}{x - 2}$

tangent slope = $4x$

$4x = \frac{2x^2}{x-2}$

$4x^2 - 8x = 2x^2$

$2x^2 - 8x = 0$

$2x(x - 4) = 0$

$x = 0$ , $x = 4$

point on the curve ... (0,3) , slope = 0
the tangent line that passes through (2,3) is y = 3

point on the curve ... (4,35), slope = 16
the tangent line that passes through (2,3) is y = 16x - 29