The gradient of a curve is ax+b at all points, where a and b are constants. Find the equation of the curve, given that it passes through the points (0,4) and (1,3), and that the tangent at (1,3) is parallel to the line y=x.
Let’s say the equation of the curve is y=f(x), so y′=ax+b. This means that y=∫ax+bdx=ax22+bx+c for some constant of integration, c. (This is a consequence of the fundamental theorem of calculus.)
As the curve passes through the point (0,4) we know that a×022+b×0+c=4⟹c=4, and as it passes through the point (1,3), we also have that a×122+b×1+c=3⟹a2+b+4=3⟹a+2b=−2. Now the tangent to the curve at (1,3) is parallel to the line y=x, which has gradient 1, so a+b=4 (since y′=ax+b).
We therefore have simultaneous equations in a and b, which are a+2b=−2,a+b=1.Subtracting, we have b=−3, and hence a=4. The equation of the curve is therefore y=2x2−3x+4.