Where a line meets a curve questions with answers

Questions

1. Question 1The number of meeting points

   Find the number of points at which the line $y = -x + 7$ meets the curve $y = x^2 + x + 6$.

   Show the worked solution

   1. Step 1: Where they meet, the y-values are equal: make one quadratic$x^2 + x + 6$ $=$ $-x + 7$$x^2 + 2x - 1 = 0$
   2. Step 2: The discriminant tells you how many times they meet$b^2 - 4ac = (2)^2 - 4(1)(-1)$$= 8$Positive: two points.
2. Question 2Tangent: find k

   The line $y = -x + k$ is a tangent to the curve $y = x^2 + 3x + 5$. Find the value of $k$.

   Show the worked solution

   1. Step 1: Where they meet, the y-values are equal$x^2 + 3x + 5$ $=$ $-x + k$$x^2 + 4x + 5 - k = 0$
   2. Step 2: Which condition on the discriminant gives a tangent (the line touches the curve once)?$b^2 - 4ac = 0$
   3. Step 3: Substitute a, b and c, expand and solve$(4)^2 - 4(1)(5 - k) = 0$$16 - 20 + 4k = 0$$4k = 4$$k = 1$
3. Question 3Where they meet

   Find the $x$-coordinates of the points where the line $y = 2x + 1$ meets the curve $y = x^2 - x - 3$.

   Show the worked solution

   1. Step 1: Where they meet, the y-values are equal: make one quadratic$x^2 - x - 3$ $=$ $2x + 1$$x^2 - 3x - 4 = 0$
   2. Step 2: Factorise and solve$(x + 1)(x - 4) = 0$$x = -1$ or $x = 4$
4. Question 4Curves with xy or y² terms

   The line $y = 3x - 1$ meets the curve $2x^2 - xy = -6$ at two points. Find the $x$-coordinates of these points.

   Show the worked solution

   1. Step 1: Substitute the line into the curve: replace every y$2x^2 - x(3x - 1) = -6$
   2. Step 2: Expand, collect every term on one side: one quadratic in x$-x^{2} + x + 6 = 0$$x^{2} - x - 6 = 0$
   3. Step 3: Factorise and solve$(x + 2)(x - 3) = 0$$x = -2$ or $x = 3$