Use the point (0, -12) from the table and substitute x=0 and y=-12 into the relation y = px^2 - 5x + q.

Question

Use the point (0, -12) from the table and substitute x=0 and y=-12 into the relation y = px^2 - 5x + q.

Asked on April 25, 2026Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

Accepted Answer · 2026-04-25T04:43:13.733Z

a) (i) Use the table to find the values of p and q. Step 1: Use the point $(0, -12)$ from the table and substitute $x=0$ and $y=-12$ into the relation $y = px^2 - 5x + q$. $$ -12 = p(0)^2 - 5(0) + q $$ $$ -12 = 0 - 0 + q $$ $$ q = -12 $$ Step 2: Use another point from the table, for example $(4, 0)$, and substitute $x=4$, $y=0$, and $q=-12$ into the relation $y = px^2 - 5x + q$. $$ 0 = p(4)^2 - 5(4) + (-12) $$ $$ 0 = 16p - 20 - 12 $$ $$ 0 = 16p - 32 $$ $$ 16p = 32 $$ $$ p = \frac{32}{16} $$ $$ p = 2 $$ The values are $p=2$ and $q=-12$. The relation is $y = 2x^2 - 5x - 12$. (ii) Copy and complete the table. Using the relation $y = 2x^2 - 5x - 12$, we calculate the missing y-values: For $x=-1$: $$ y = 2(-1)^2 - 5(-1) - 12 $$ $$ y = 2(1) + 5 - 12 $$ $$ y = 2 + 5 - 12 $$ $$ y = 7 - 12 = -5 $$ For $x=1$: $$ y = 2(1)^2 - 5(1) - 12 $$ $$ y = 2(1) - 5 - 12 $$ $$ y = 2 - 5 - 12 $$ $$ y = -3 - 12 = -15 $$ For $x=2$: $$ y = 2(2)^2 - 5(2) - 12 $$ $$ y = 2(4) - 10 - 12 $$ $$ y = 8 - 10 - 12 $$ $$ y = -2 - 12 = -14 $$ For $x=3$: $$ y = 2(3)^2 - 5(3) - 12 $$ $$ y = 2(9) - 15 - 12 $$ $$ y = 18 - 15 - 12 $$ $$ y = 3 - 12 = -9 $$ The completed table is: | x | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 | |---|----|----|----|---|---|---|---|---|---| | y | 21 | 6 | -5 | -12 | -15 | -14 | -9 | 0 | 13 | b) Using scales of 2cm to 1 unit on the x-axis and 2 cm to 5 units on the y-axis, draw the graph of the relation for $-3 \le x \le 5$. Step 1: Draw the x-axis and y-axis on graph paper. Step 2: Mark the x-axis from -3 to 5, with each 2 cm representing 1 unit. Step 3: Mark the y-axis to cover the range of y-values (from -15 to 21), with each 2 cm representing 5 units. Step 4: Plot the points from the completed table: $(-3, 21)$, $(-2, 6)$, $(-1, -5)$, $(0, -12)$, $(1, -15)$, $(2, -14)$, $(3, -9)$, $(4, 0)$, $(5, 13)$. Step 5: Draw a smooth curve connecting all the plotted points. This curve represents the graph of $y = 2x^2 - 5x - 12$. c) Use the graph to find: (i) y when x = 1.8 To find $y$ when $x=1.8$ from the graph, locate $x=1.8$ on the x-axis. Move vertically from $x=1.8$ to intersect the curve. From the intersection point, move horizontally to the y-axis to read the corresponding y-value. Using the equation $y = 2x^2 - 5x - 12$: $$ y = 2(1.8)^2 - 5(1.8) - 12 $$ $$ y = 2(3.24) - 9 - 12 $$ $$ y = 6.48 - 9 - 12 $$ $$ y = 6.48 - 21 $$ $$ y = -14.52 $$ From the graph, $y$ would be approximately $\boxed{-14.5}$. (ii) x when y = -8 To find $x$ when $y=-8$ from the graph, locate $y=-8$ on the y-axis. Move horizontally from $y=-8$ to intersect the curve. From the intersection points, move vertically to the x-axis to read the corresponding x-values. Using the equation $y = 2x^2 - 5x - 12$: $$ -8 = 2x^2 - 5x - 12 $$ $$ 0 = 2x^2 - 5x - 4 $$ Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $$ x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(-4)}}{2(2)} $$ $$ x = \frac{5 \pm \sqrt{25 + 32}}{4} $$ $$ x = \frac{5 \pm \sqrt{57}}{4} $$ $$ x_1 = \frac{5 + \sqrt{57}}{4} \approx \frac{5 + 7.5498}{4} \approx \frac{12.5498}{4} \approx 3.137 $$ $$ x_2 = \frac{5 - \sqrt{57}}{4} \approx \frac{5 - 7.5498}{4} \approx \frac{-2.5498}{4} \approx -0.637 $$ From the graph, $x$ would be approximately $\boxed{3.1 \text{ or } -0.6}$. That's 2 down. 3 left today — send the next one.

Accepted Answer · 2026-04-25T04:43:13.733Z

a) (i) Use the table to find the values of p and q. Step 1: Use the point $(0, -12)$ from the table and substitute $x=0$ and $y=-12$ into the relation $y = px^2 - 5x + q$. $$ -12 = p(0)^2 - 5(0) + q $$ $$ -12 = 0 - 0 + q $$ $$ q = -12 $$ Step 2: Use another point from the table, for example $(4, 0)$, and substitute $x=4$, $y=0$, and $q=-12$ into the relation $y = px^2 - 5x + q$. $$ 0 = p(4)^2 - 5(4) + (-12) $$ $$ 0 = 16p - 20 - 12 $$ $$ 0 = 16p - 32 $$ $$ 16p = 32 $$ $$ p = \frac{32}{16} $$ $$ p = 2 $$ The values are $p=2$ and $q=-12$. The relation is $y = 2x^2 - 5x - 12$. (ii) Copy and complete the table. Using the relation $y = 2x^2 - 5x - 12$, we calculate the missing y-values: For $x=-1$: $$ y = 2(-1)^2 - 5(-1) - 12 $$ $$ y = 2(1) + 5 - 12 $$ $$ y = 2 + 5 - 12 $$ $$ y = 7 - 12 = -5 $$ For $x=1$: $$ y = 2(1)^2 - 5(1) - 12 $$ $$ y = 2(1) - 5 - 12 $$ $$ y = 2 - 5 - 12 $$ $$ y = -3 - 12 = -15 $$ For $x=2$: $$ y = 2(2)^2 - 5(2) - 12 $$ $$ y = 2(4) - 10 - 12 $$ $$ y = 8 - 10 - 12 $$ $$ y = -2 - 12 = -14 $$ For $x=3$: $$ y = 2(3)^2 - 5(3) - 12 $$ $$ y = 2(9) - 15 - 12 $$ $$ y = 18 - 15 - 12 $$ $$ y = 3 - 12 = -9 $$ The completed table is: | x | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 | |---|----|----|----|---|---|---|---|---|---| | y | 21 | 6 | -5 | -12 | -15 | -14 | -9 | 0 | 13 | b) Using scales of 2cm to 1 unit on the x-axis and 2 cm to 5 units on the y-axis, draw the graph of the relation for $-3 \le x \le 5$. Step 1: Draw the x-axis and y-axis on graph paper. Step 2: Mark the x-axis from -3 to 5, with each 2 cm representing 1 unit. Step 3: Mark the y-axis to cover the range of y-values (from -15 to 21), with each 2 cm representing 5 units. Step 4: Plot the points from the completed table: $(-3, 21)$, $(-2, 6)$, $(-1, -5)$, $(0, -12)$, $(1, -15)$, $(2, -14)$, $(3, -9)$, $(4, 0)$, $(5, 13)$. Step 5: Draw a smooth curve connecting all the plotted points. This curve represents the graph of $y = 2x^2 - 5x - 12$. c) Use the graph to find: (i) y when x = 1.8 To find $y$ when $x=1.8$ from the graph, locate $x=1.8$ on the x-axis. Move vertically from $x=1.8$ to intersect the curve. From the intersection point, move horizontally to the y-axis to read the corresponding y-value. Using the equation $y = 2x^2 - 5x - 12$: $$ y = 2(1.8)^2 - 5(1.8) - 12 $$ $$ y = 2(3.24) - 9 - 12 $$ $$ y = 6.48 - 9 - 12 $$ $$ y = 6.48 - 21 $$ $$ y = -14.52 $$ From the graph, $y$ would be approximately $\boxed{-14.5}$. (ii) x when y = -8 To find $x$ when $y=-8$ from the graph, locate $y=-8$ on the y-axis. Move horizontally from $y=-8$ to intersect the curve. From the intersection points, move vertically to the x-axis to read the corresponding x-values. Using the equation $y = 2x^2 - 5x - 12$: $$ -8 = 2x^2 - 5x - 12 $$ $$ 0 = 2x^2 - 5x - 4 $$ Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $$ x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(-4)}}{2(2)} $$ $$ x = \frac{5 \pm \sqrt{25 + 32}}{4} $$ $$ x = \frac{5 \pm \sqrt{57}}{4} $$ $$ x_1 = \frac{5 + \sqrt{57}}{4} \approx \frac{5 + 7.5498}{4} \approx \frac{12.5498}{4} \approx 3.137 $$ $$ x_2 = \frac{5 - \sqrt{57}}{4} \approx \frac{5 - 7.5498}{4} \approx \frac{-2.5498}{4} \approx -0.637 $$ From the graph, $x$ would be approximately $\boxed{3.1 \text{ or } -0.6}$. That's 2 down. 3 left today — send the next one.

Use the point (0, -12) from the table and substitute x=0 and y=-12 into the relation y = px^2 - 5x + q.