Find the tangent to the curve y=(3/4)x^2 -4x^(1/2) +7 at x=4, expressing it in the form ax+by+c=0.

In general, the equation of a line is y-y0=m(x-x0), where m is the gradient and (x0,y0) is a point on the line. We need to find the gradient and a point on our tangent. Step 1: Find gradient. Since our tangent is at x=4 of the curve, we know it has the same gradient as the curve at x=4. dy/dx= (3/2)x-2x^(-1/2). Therefore, by substituting x=4 into dy/dx we see that the gradient at x=4 is (3/2)(4)-2(4)^(-1/2)=5 Step 2: Find (x0,y0)As the tangent passes the curve at x=4, it must contain the corresponding y coordinate at x=4. Set x=4 into equation of curve and we obtain y= =(3/4)(4)^2 -4(4)^(1/2) +7 = 11. So (x0,y0) = (4,11) Step 3: Equation of tangent. Set the above into y-y0 = m(x-x0) à y-11=5(x-4) By expanding we then find that the equation of the tangent is 5x-y-9=0 where a=5, b=-1 and c=-9