how do I get the tangent coordinates if I am given the quadratic function and the slope(gradient)

Question

I have the following task: calculate the coordinates of the intersections point between a straight line with a given slope and a quadratic function, so that you only receive one intersection instead of the normal two or none.
I am given the slope gradient m and the quadratic equation.
In this example its

y=x^2+3x-2   m=1

Answer 1

Let the equation of straight line be $y = mx +c$. Solve for the intersection: $$ mx+c = x^2 +3x -2$$ You get a quadratic in x. $$ x^2 + (3-m) x -(2+c) = 0$$

To have only one intersection point, there should be only value of $x$ satisfying the above equation. So you make the discriminant of the above equation zero.

So now your x is simply $(m-3)/2$.

You know the value of $m$, so you know $x$, then you can find $y$ to know the intersection point

Answer 2

First, find the derivative of $y = x^2 + 3x - 2$.

$$y' = 2x + 3$$

Now set $2x+3 = m = 1$.

Solve for $x$ to find the $x$ coordinate of the point you need.

Then place your solution for $x$ into the equation $y = x^2 + 3x -2$ to find the y-coordinate.

Answer 3

You don’t really need calculus to solve this problem thanks to a property of the parabola, namely, that a tangent to it doesn’t intersect the parabola at any other points. So, you are essentially looking for values of $b$ such that the line $y=x+b$ and the parabola $y=x^2+3x-2$ intersect in a single point. Setting the right-hand sides equal to each other and rearranging, you get the quadratic equation $$x^2+2x-(b+2)=0.$$ This equation has a single solution when its discriminant is zero, i.e., when $$4-4(b+2)=0.$$ I hope you can take it from here.