I'm looking for other coordinates on the curve where the tangent is parallel to y=bx+3. I have solved for a, b, c and tangent equation.

Question. (-3,a) is on curve y=5x-7/2x+10. Let us assume you mean $\displaystyle y=5x-\frac{7}{2x}+10$ . Not a good idea to be cheap with parentheses... Thus, $\displaystyle y(-3)=-15+\frac{7}{6}+10=-\frac{23}{6}=a$ The tangent at (-3,a) is parallel to y=bx+3 and cuts the y-axis at (0,c). The tangent at $\displaystyle (-3,-23\slash 6)$ has a slope equal to $\displaystyle y'(-3)$ , and $\displaystyle y'=5+\frac{7}{2x^2}\Longrightarrow y'(-3)=$$\displaystyle 5+\frac{7}{18}=\frac{97}{18}\Longrightarrow b=\frac{97}{18}$
The number we've been getting are nasty so perhaps that is not the function you meant. Check this.

Tonio[/color]

How do I get the __other coordinates__ on the curve where the tangent is parallel to y=bx+3? I’m sure I need to use the gradient perpendicular to the tangent (m=-1/4) and set it equal to y=5x-7/2x+10 or y`equation and solve for x? Any help would be appreciated.
Workings

I have solved a,b,c (-5.5,4,-6.5) and the tangent equation y=4x-6.5. But graphing the curve shows me there is 2 hyperbolas so I need the points on the curve for the other tangent equation. 1. st x = -3 into y=(5x-7)/(2x+10) to get -5.5=a ;. (-3,-5.5). 2. find y` of y=(5x-7)/(2x+10). Use quotient rule..vu`-uv`/v^2. Then you get [5(2x+10)-2(5x-7)]/[(2x+10)^2]. Then st x =-3..y`=4. 3. y`=4 is the gradient :. b=4 in y=bx+3. 4. Find equation for tangent at (-3,-5.5,m=4) using y-y1=m(x-x1). Becomes y=4x-6.5. As y=mx+c:….c=-6.5. Hence A=-5.5, b=4, c=-6.5 and tangent equation is y=4x-6.5