Graph features questions with answers

Questions

1. Question 1Where it crosses the y-axis

   Find the coordinates of the point where the curve $y = 2x^{2} + 3x + 4$ 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 + 3(0) + 4$$\phantom{y} = 4$The curve crosses the $y$-axis at $(0, 4)$.
2. Question 2x-intercepts from brackets

   Find the $x$-coordinates of the points where the curve $y = (x + 2)(x + 5)$ meets the $x$-axis.

   Show the worked solution

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

   Find the $x$-coordinates of the points where the curve $y = x^{2} - 3x - 10$ meets the $x$-axis.

   Show the worked solution

   1. Step 1: Which coordinate is 0 on the x-axis?On the $x$-axis, $y = 0$.$x^{2} - 3x - 10 = 0$
   2. Step 2: Factorise$(x - 5)(x + 2) = 0$
   3. Step 3: Each bracket can be zero$x = 5$ or $x = -2$
4. Question 4Line of symmetry

   Find the equation of the line of symmetry of the curve $y = (x - 4)(x - 1)$.

   Show the worked solution

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

   Find the coordinates of the turning point of the curve $y = -x^2 - 1$.

   Show the worked solution

   1. Step 1: Is the turning point a maximum or a minimum?The square has a minus in front, so it is a maximum point.
   2. Step 2: Make the square 0 to find x, then find yA square is never negative: $x^2 = 0$ when $x = 0$, and with the minus in front y is greatest there.Then $y = -(0)^2 - 1$$y = -1$Turning point: $(0, -1)$
6. Question 6Turning point: complete the square

   Find the coordinates of the turning point of the curve $y = x^{2} + 4x - 4$.

   Show the worked solution

   1. Step 1: Complete the square$x^{2} + 4x - 4 = (x + 2)^2 - 4 - 4$$= (x + 2)^2 - 8$
   2. Step 2: Make the bracket 0 to find x, then find y$x = -2$, so $y = -8$Turning point: $(-2, -8)$
7. Question 7Maximum or minimum value

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

   Show the worked solution

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