Gradient Function Query (1 Viewer)

[Finx]

Hey all =]

Just seeking confirmation on this..

Find the coordinates of the point on the parabola y = 6x²-13x+5 at which the tangent is parallel to the line y+x=4.

My working:
Equation of line: y = -x + 4
.'. gradient of line = -1
dy/dx of parabola = 12x-13
-1 = 12x - 13
12 = 12x
x = 1
sub x into parabola: 6(1)²-13(1)+5 = -2
coordinates of point: (1,-2)

Answer sheet working:
All of the above, except for the underlined, which it has:
sub x into dy/dx of parabola: 12(1)-13 = -1
coordinates of point: (1,-1)

I'm wondering which is correct - substituing the x value into the parabola or substituting into the differentiation?

Thanks =D

 

[Dr. Zoidberg]

Into parabola, so the first one.

When you sub into dy/dx, you are just finding the rate change of the parabola at that point

 

[piekarz]

Sup.

Well, in my opinion, you have to substitue the value of X in the parabola.

Theoretically, if given a point of that parabola, it should satisfy the equation.

Therefore, from y = 6x²-13x+5

(from the answers sheet), the point is (1, -1).

therefore: -1 = 6 - 13 + 5
-1 = -2? No.

I would say you did it correctly when the point was (1, -2)

Therefore: y = 6x²-13x+5
-2 = 6 - 13 + 5
-2 = -2

In my view, thats correct, but i only got 76% in the last extension test. So meh.

OMFG DR. ZOIDBERG ANSWERED 1 MINUTE BEFORE ME WTF!

I wanted to answer ur Q first brooo O well

 

[Aplus]

You substitute it into the original equation. The original equation is the one which tells you the points of the co-ordinates which lie on the graph. If you substitute it in to the first derivative, you're just going to get the gradient of the tangent at that point.

 

[Finx]

Doth this mean that the answer sheet speaketh of lies?! D=

 

[Aplus]

Differentiating the equation of the parabola y = 6x² - 13x + 5 gives:
dy/dx = 12x - 13 (As the first derivative, this is informs us of the rate of change as one progresses through the curve)

If the tangent is parallel to the line y + x = 4, then they must share the same gradient, as m₁ = m₂ for parallel lines.
The gradient of the line y + x = 4 is -1

Therefore, the gradient of the tangent is -1
Substituting this into the gradient function:
12x - 13 = -1
12x = 12
x = 1

This informs us that when the x co-ordinate of the parabola is at 1, a tangent will occur. We now need to calculate at what exact point this would occur. To do this, we substitute the value of x = 1 into the original equation of the parabola y = 6x² - 13x + 5:
y = 6(1)² - 13(1) + 5
= -2

Thus, the co-ordinates of the point on the parabola y = 6x²-13x+5 at which the tangent is parallel to the line y+x=4, is (1,-2)

 

[Finx]

This informs us that when the x co-ordinate of the parabola is at 1, a tangent will occur.

I thought it meant that at that x co-ordinate (in this case, 1), the tangent to that point will have a gradient of the substituted value (in this case, -1)

Am I right, or am I just misunderstanding your explanation?

 

[tommykins]

Hey all =]

Just seeking confirmation on this..

Find the coordinates of the point on the parabola y = 6x²-13x+5 at which the tangent is parallel to the line y+x=4.

My working:
Equation of line: y = -x + 4
.'. gradient of line = -1
dy/dx of parabola = 12x-13
-1 = 12x - 13
12 = 12x
x = 1
sub x into parabola: 6(1)²-13(1)+5 = -2
coordinates of point: (1,-2)

Answer sheet working:
All of the above, except for the underlined, which it has:
sub x into dy/dx of parabola: 12(1)-13 = -1
coordinates of point: (1,-1)

I'm wondering which is correct - substituing the x value into the parabola or substituting into the differentiation?

Thanks =D

You sub x = 1 into the equation of the parabola to find the y co-ordinate of the point on the parabola when x = 1.

 

[Aplus]

I thought it meant that at that x co-ordinate (in this case, 1), the tangent to that point will have a gradient of the substituted value (in this case, -1)

Am I right, or am I just misunderstanding your explanation?

I should probably have been more specific. The tangent which is parallel to the aforementioned line.

 

[Shoom]

Is this 2u? If so im even more screwed for tomorrows test.

 

[Finx]

I should probably have been more specific. The tangent which is parallel to the aforementioned line.

Sweet, makes perfect sense, ta =]

Is this 2u? If so im even more screwed for tomorrows test.

Yeah this is from a 2U calculus school paper.