Table of values for a quadratic questions with answers

Questions

1. Question 1y = x² + bx + c

   Complete the table of values for $y = x^{2} - 4x - 1$.

   Show the worked solution

   1. Step 1: Put each missing x into the equation and work out each term. Square a NEGATIVE x first: (−2)² = 4, not −4When $x = -2$:$y = (-2)^2 - 4(-2) - 1$$= 4 + 8 - 1$$= 11$When $x = 1$:$y = (1)^2 - 4(1) - 1$$= 1 - 4 - 1$$= -4$When $x = 3$:$y = (3)^2 - 4(3) - 1$$= 9 - 12 - 1$$= -4$When $x = 4$:$y = (4)^2 - 4(4) - 1$$= 16 - 16 - 1$$= -1$
   2. Step 2: Check: the y-values are symmetrical about the line of symmetryThe $y$-values are the same for $x = 0$ and $x = 4$, and so on: they are symmetrical about $x = 2$.
2. Question 2y = x² + bx + c

   Complete the table of values for $y = x^{2} + 2x + 5$.

   Show the worked solution

   1. Step 1: Put each missing x into the equation and work out each term. Square a NEGATIVE x first: (−2)² = 4, not −4When $x = -1$:$y = (-1)^2 + 2(-1) + 5$$= 1 - 2 + 5$$= 4$When $x = 1$:$y = (1)^2 + 2(1) + 5$$= 1 + 2 + 5$$= 8$When $x = 2$:$y = (2)^2 + 2(2) + 5$$= 4 + 4 + 5$$= 13$When $x = 4$:$y = (4)^2 + 2(4) + 5$$= 16 + 8 + 5$$= 29$
   2. Step 2: Check: the y-values are symmetrical about the line of symmetryThe $y$-values are the same for $x = -3$ and $x = 1$, and so on: they are symmetrical about $x = -1$.
3. Question 3y = −x² + bx + c

   Complete the table of values for $y = -x^{2} + 2x + 6$.

   Show the worked solution

   1. Step 1: Put each missing x into the equation and work out each term. Square a NEGATIVE x first: (−2)² = 4, not −4; and −x² means −(x²)When $x = -4$:$y = -(-4)^2 + 2(-4) + 6$$= -16 - 8 + 6$$= -18$When $x = -1$:$y = -(-1)^2 + 2(-1) + 6$$= -1 - 2 + 6$$= 3$When $x = 3$:$y = -(3)^2 + 2(3) + 6$$= -9 + 6 + 6$$= 3$
   2. Step 2: Check: the y-values are symmetrical about the line of symmetryThe $y$-values are the same for $x = -1$ and $x = 3$, and so on: they are symmetrical about $x = 1$.
4. Question 4y = −x² + bx + c

   Complete the table of values for $y = -x^{2} + 3x - 2$.

   Show the worked solution

   1. Step 1: Put each missing x into the equation and work out each term. Square a NEGATIVE x first: (−2)² = 4, not −4; and −x² means −(x²)When $x = -3$:$y = -(-3)^2 + 3(-3) - 2$$= -9 - 9 - 2$$= -20$When $x = 0$:$y = -(0)^2 + 3(0) - 2$$= 0 - 2$$= -2$When $x = 4$:$y = -(4)^2 + 3(4) - 2$$= -16 + 12 - 2$$= -6$
   2. Step 2: Check: the y-values are symmetrical about the line of symmetryThe $y$-values are the same for $x = -1$ and $x = 4$, and so on: they are symmetrical about $x = 1.5$.
5. Question 5y = 2x² + bx + c and y = −2x² + bx + c

   Complete the table of values for $y = 2x^{2} - 2x - 2$.

   Show the worked solution

   1. Step 1: Put each missing x into the equation and work out each term. Square a NEGATIVE x first: (−2)² = 4, not −4When $x = -2$:$y = 2(-2)^2 - 2(-2) - 2$$= 8 + 4 - 2$$= 10$When $x = -1$:$y = 2(-1)^2 - 2(-1) - 2$$= 2 + 2 - 2$$= 2$When $x = 0$:$y = 2(0)^2 - 2(0) - 2$$= 0 - 2$$= -2$When $x = 4$:$y = 2(4)^2 - 2(4) - 2$$= 32 - 8 - 2$$= 22$
   2. Step 2: Check: the y-values are symmetrical about the line of symmetryThe $y$-values are the same for $x = -3$ and $x = 4$, and so on: they are symmetrical about $x = 0.5$.
6. Question 6y = 2x² + bx + c and y = −2x² + bx + c

   Complete the table of values for $y = 2x^{2} - 4$.

   Show the worked solution

   1. Step 1: Put each missing x into the equation and work out each term. Square a NEGATIVE x first: (−2)² = 4, not −4When $x = -2$:$y = 2(-2)^2 - 4$$= 8 - 4$$= 4$When $x = -1$:$y = 2(-1)^2 - 4$$= 2 - 4$$= -2$When $x = 3$:$y = 2(3)^2 - 4$$= 18 - 4$$= 14$
   2. Step 2: Check: the y-values are symmetrical about the line of symmetryThe $y$-values are the same for $x = -2$ and $x = 2$, and so on: they are symmetrical about $x = 0$.