Graph features questions with answers

Questions

1. Question 1Where it crosses the y-axis

   Find the coordinates of the point where the curve $y = -2x^{2} - 6x - 7$ crosses the $y$-axis.

   Show the worked solution

   1. Step 1: Which coordinate is 0 on the y-axis?On the $y$-axis, $x = 0$.
   2. Step 2: Substitute, then write the point$y = -2(0)^2 - 6(0) - 7$$\phantom{y} = -7$The curve crosses the $y$-axis at $(0, -7)$.
2. Question 2x-intercepts from brackets

   Find the $x$-coordinates of the points where the curve $y = -(x + 3)(x + 5)$ meets the $x$-axis.

   Show the worked solution

   1. Step 1: Which coordinate is 0 on the x-axis?On the $x$-axis, $y = 0$.$-(x + 3)(x + 5) = 0$
   2. Step 2: Each bracket can be zero$x = -3$ or $x = -5$
3. Question 3x-intercepts: factorise first

   Find the $x$-coordinates of the points where the curve $y = x^{2} - 4x - 5$ meets the $x$-axis.

   Show the worked solution

   1. Step 1: Which coordinate is 0 on the x-axis?On the $x$-axis, $y = 0$.$x^{2} - 4x - 5 = 0$
   2. Step 2: Factorise$(x - 5)(x + 1) = 0$
   3. Step 3: Each bracket can be zero$x = 5$ or $x = -1$
4. Question 4Line of symmetry

   Find the equation of the line of symmetry of the curve $y = -(x + 2)(x + 4)$.

   Show the worked solution

   1. Step 1: Which coordinate is 0 on the x-axis?On the $x$-axis, $y = 0$.$-(x + 2)(x + 4) = 0$
   2. Step 2: Each bracket can be zero$x = -2$ or $x = -4$
   3. Step 3: Find the line of symmetry from the x-intercepts$x = \dfrac{-2 + (-4)}{2}$$x = -3$
5. Question 5Turning point of ±(x − h)² + k

   Find the coordinates of the turning point of the curve $y = (x - 4)^2 + 6$.

   Show the worked solution

   1. Step 1: Is the turning point a maximum or a minimum?The square is added, so it is a minimum point.
   2. Step 2: Make the square 0 to find x, then find yA square is never negative: $(x - 4)^2 = 0$ when $x = 4$, so y is smallest there.Then $y = (0)^2 + 6$$y = 6$Turning point: $(4, 6)$
6. Question 6Maximum or minimum value

   State the minimum value of $y$ for the curve $y = (x + 1)^2 - 4$.

   Show the worked solution

   1. Step 1: Find the x that makes the square 0A square is never negative: $(x + 1)^2 = 0$ when $x = -1$, so y is smallest there.
   2. Step 2: Find y thereThen $y = (0)^2 - 4$$y = -4$The minimum value of $y$ is $-4$.