Could someone help me with this calculus question???

Question

Find the equation that passes through (2, 3) if its slope is given by:

dy/dx= 4x-3

Answer 1

m = 4x2 - 3 = 5
y - 3 = 5(x - 2)
y = 5x -7

Answer 2

It cannot be a straight line if the slope varies with x. A straight line would have constant slope.

If dy/dx= 4x-3 then

y = 2x^2 -3x plus an arbitrary constant C. Find the value of that constant so (2,3) will be on the curve;

y = 2x^2 - 3x + C
3 = 2(2^2) - 3(2) + C = 2 +C

So C must equal 1

The curve is given by

y = 2x^2 - 3x + 1

Answer 3

since slope is 4x-3
and it passes through 2,3
we get 4(2)-3=5
slope = 5 =m
y=mx+b
3=5(2)+b
3=10+b
b=-7
y=5x-7 (done)