How do you determine q in this equation?

Answer

Dear Sara

It depends on what your prior knowledge is. I see you’re 14, so you probably haven’t seen any derivatives yet. How did you define the term “tangent”? Like a line that intersects the parabola in just one point? Let’s say yes. Then you find the intersections of that parabola with that line, and you express that there is only one solution. Equating the y-values ​​yields 4x+q=4x²-12x+8, or 4x²-16x+(8-q)=0. Expressing that this equation has only one solution is equivalent to equating the discriminant to zero. So 16²-4.4(8-q)=0, which gives q=-8, and then x=16/8=2. Substituting this value for x in the equation of the parabola gives the corresponding y-value 0. So the coordinate of the tangent point is (2,0).

If you have derivatives, you know that the slope of the tangent line is 4 (the coefficient of x in the equation y=4x+q), so the derivative at the tangent point must be 4. The derived function is y’=8x-12, which yields 4 for x=2. This gives as in previous section (2.0) for the tangent point, and the equation of the tangent line is (y-0)=4(x-2), yielding q=-8.

I hope this helps!

Regards

—Hendrik Van Maldeghem