A curve has equation $y=5-2x+x^2$ and a line has equation $y=2x+k$, where $k$ is a constant. Answer the question in the body of text below.

Question

For one value of $k$, the line intersects the curve at two points, A and B, where the co-ordinates of A are $(-2,13)$.

Can you help me find the co-ordinates of B?

I don't remember how curve co-ordinate equations work, particularly equations like the above, can someone please tell me how they work?

Answer 1

The hint.

Since $$13=-4+k,$$ we obtain $k=17$ and solve the following equation: $$x^2-2x+5=2x+17.$$ I got $B(6,29)$.

Answer 2

Hint: Solve the quadratic $$2x+k=5-2x+x^2$$ for $x$

Answer 3

$$5-2x+x^2=2x+k$$

it will give you point $P(x_1,y_1)$ and $Q(x_2,y_2)$ where the line cuts the curve but you already have a point(-2,13) thus $-2$ satisfies this equation

$$5-2(-2)+(-2)^2=2(-2)+k$$ $$5+4+4=-4+k$$ $$k=17$$

now your equation becomes $$5-2x+x^2=2x+17$$ $$x^2-4x-12=0$$ sum of two roots is $-2+\beta=\frac{4}{1}$ $$b=6$$

thus your point B is $(6,5-2(6)+6^2) $ which his B$(6,29)$