Calculus Question (1 Viewer)

[Resource]

Find the coordinates of point Q where the tangent to the curve y=(5x^2)-3x is parallel to the line 7x-y+3=0.

I keep getting the answer Point Q: (-2/5, 2) but the BOB says the answer is (1,2)... What am I doing wrong?

 

[Trebla]

Find the coordinates of point Q where the tangent to the curve y=(5x^2)-3x is parallel to the line 7x-y+3=0.

I keep getting the answer Point Q: (-2/5, 2) but the BOB says the answer is (1,2)... What am I doing wrong?

I'm assuming Q is the point of contact of the tangent. Taking the derivative:
dy/dx = 10x - 3
We want to find the x value which causes to the gradient of the tangent to be 7 since the equation of the line that it needs to be parallel to is y = 7x + 3
10x - 3 = 7
x = 1
When x = 1, y = 2 when substituting into the curve.

 

[Resource]

Oh thank you I understand where I went wrong now.

 

[Resource]

I have another question that I'm stuck on guys:

Differentiate: 3/[4(2x+7)^9]

I could do it if it didn't have the 3/4 in it but I'm not sure what to do with it.

 

[braintic]

I have another question that I'm stuck on guys:

Differentiate: 3/[4(2x+7)^9]

I could do it if it didn't have the 3/4 in it but I'm not sure what to do with it.

The answer is just 3/4 of the answer you would get without the 3/4.