Answer :
a)(i) Use the table to find the values of [tex]\( p \)[/tex] and [tex]\( q \)[/tex]
Given the quadratic relation:
[tex]\[ y = p x^2 - 5 x + q \][/tex]
We are provided with points [tex]\((-3, 21)\)[/tex], [tex]\((-2, 6)\)[/tex], [tex]\((0, -12)\)[/tex], [tex]\((4, 0)\)[/tex], and [tex]\((5, 13)\)[/tex].
Using the points [tex]\((-3, 21)\)[/tex] and [tex]\((-2, 6)\)[/tex]:
1. Substitute [tex]\((x, y) = (-3, 21)\)[/tex] into the equation:
[tex]\[ 21 = p (-3)^2 - 5 (-3) + q \][/tex]
[tex]\[ 21 = 9p + 15 + q \][/tex]
[tex]\[ 21 = 9p + q + 15 \][/tex]
[tex]\[ 6 = 9p + q \][/tex]
2. Substitute [tex]\((x, y) = (-2, 6)\)[/tex] into the equation:
[tex]\[ 6 = p (-2)^2 - 5 (-2) + q \][/tex]
[tex]\[ 6 = 4p + 10 + q \][/tex]
[tex]\[ 6 = 4p + q + 10 \][/tex]
[tex]\[ -4 = 4p + q \][/tex]
Solving the system of equations:
[tex]\[ 6 = 9p + q \][/tex]
[tex]\[ -4 = 4p + q \][/tex]
Subtract the second equation from the first:
[tex]\[ (6 - (-4)) = (9p + q) - (4p + q) \][/tex]
[tex]\[ 10 = 5p \][/tex]
[tex]\[ p = 2 \][/tex]
Now substitute [tex]\( p = 2 \)[/tex] into either equation to find [tex]\( q \)[/tex]:
[tex]\[ 6 = 9 \cdot 2 + q \][/tex]
[tex]\[ 6 = 18 + q \][/tex]
[tex]\[ q = 6 - 18 \][/tex]
[tex]\[ q = -12 \][/tex]
Thus, the values are:
[tex]\[ p = 2 \][/tex]
[tex]\[ q = -12 \][/tex]
a)(ii) Copy and complete the table
Using [tex]\( p = 2 \)[/tex] and [tex]\( q = -12 \)[/tex], fill in the values of [tex]\( y \)[/tex]:
1. For [tex]\( x = -1 \)[/tex]:
[tex]\[ y = 2(-1)^2 - 5(-1) - 12 \][/tex]
[tex]\[ y = 2(1) + 5 - 12 \][/tex]
[tex]\[ y = 2 + 5 - 12 \][/tex]
[tex]\[ y = -5 \][/tex]
2. For [tex]\( x = 1 \)[/tex]:
[tex]\[ y = 2(1)^2 - 5(1) + q \][/tex]
[tex]\[ y = 2(1) - 5(1) - 12 \][/tex]
[tex]\[ y = 2 - 5 - 12 \][/tex]
[tex]\[ y = -15 \][/tex]
3. For [tex]\( x = 2 \)[/tex]:
[tex]\[ y = 2(2)^2 - 5(2) - 12 \][/tex]
[tex]\[ y = 2(4) - 5(2) - 12 \][/tex]
[tex]\[ y = 8 - 10 - 12 \][/tex]
[tex]\[ y = -14 \][/tex]
4. For [tex]\( x = 3 \)[/tex]:
[tex]\[ y = 2(3)^2 - 5(3) - 12 \][/tex]
[tex]\[ y = 2(9) - 5(3) - 12 \][/tex]
[tex]\[ y = 18 - 15 - 12 \][/tex]
[tex]\[ y = -9 \][/tex]
Updating the table:
[tex]\[ \begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline x & -3 & -2 & -1 & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline y & 21 & 6 & -5 & -12 & -15 & -14 & -9 & 0 & 13 \\ \hline \end{array} \][/tex]
b) Graphing the relation
Once you have completed the table, plot the points [tex]\((-3, 21)\)[/tex], [tex]\((-2, 6)\)[/tex], [tex]\((-1, -5)\)[/tex], [tex]\((0, -12)\)[/tex], [tex]\((1, -15)\)[/tex], [tex]\((2, -14)\)[/tex], [tex]\((3, -9)\)[/tex], [tex]\((4, 0)\)[/tex], and [tex]\((5, 13)\)[/tex] on a graph using the given scales.
c) Use the graph to find:
(i) [tex]\( y \)[/tex] when [tex]\( x = 1.8 \)[/tex]:
Reading the value from the graph or using the equation:
[tex]\[ y = 2(1.8)^2 - 5(1.8) - 12 \][/tex]
[tex]\[ y = 2(3.24) - 9 - 12 \][/tex]
[tex]\[ y = 6.48 - 9 - 12 \][/tex]
[tex]\[ y = -14.52 \][/tex]
Hence, [tex]\( y = -14.52 \)[/tex].
(ii) [tex]\( x \)[/tex] when [tex]\( y = -8 \)[/tex]:
Reading the value from the graph, we see that there are no [tex]\( x \)[/tex] values corresponding to [tex]\( y = -8 \)[/tex]. This indicates the curve does not intersect [tex]\( y = -8 \)[/tex] within the given range of [tex]\( x \)[/tex] values.
Thus, [tex]\( x \)[/tex] when [tex]\( y = -8 \)[/tex] has no valid solutions in the specified domain.
Given the quadratic relation:
[tex]\[ y = p x^2 - 5 x + q \][/tex]
We are provided with points [tex]\((-3, 21)\)[/tex], [tex]\((-2, 6)\)[/tex], [tex]\((0, -12)\)[/tex], [tex]\((4, 0)\)[/tex], and [tex]\((5, 13)\)[/tex].
Using the points [tex]\((-3, 21)\)[/tex] and [tex]\((-2, 6)\)[/tex]:
1. Substitute [tex]\((x, y) = (-3, 21)\)[/tex] into the equation:
[tex]\[ 21 = p (-3)^2 - 5 (-3) + q \][/tex]
[tex]\[ 21 = 9p + 15 + q \][/tex]
[tex]\[ 21 = 9p + q + 15 \][/tex]
[tex]\[ 6 = 9p + q \][/tex]
2. Substitute [tex]\((x, y) = (-2, 6)\)[/tex] into the equation:
[tex]\[ 6 = p (-2)^2 - 5 (-2) + q \][/tex]
[tex]\[ 6 = 4p + 10 + q \][/tex]
[tex]\[ 6 = 4p + q + 10 \][/tex]
[tex]\[ -4 = 4p + q \][/tex]
Solving the system of equations:
[tex]\[ 6 = 9p + q \][/tex]
[tex]\[ -4 = 4p + q \][/tex]
Subtract the second equation from the first:
[tex]\[ (6 - (-4)) = (9p + q) - (4p + q) \][/tex]
[tex]\[ 10 = 5p \][/tex]
[tex]\[ p = 2 \][/tex]
Now substitute [tex]\( p = 2 \)[/tex] into either equation to find [tex]\( q \)[/tex]:
[tex]\[ 6 = 9 \cdot 2 + q \][/tex]
[tex]\[ 6 = 18 + q \][/tex]
[tex]\[ q = 6 - 18 \][/tex]
[tex]\[ q = -12 \][/tex]
Thus, the values are:
[tex]\[ p = 2 \][/tex]
[tex]\[ q = -12 \][/tex]
a)(ii) Copy and complete the table
Using [tex]\( p = 2 \)[/tex] and [tex]\( q = -12 \)[/tex], fill in the values of [tex]\( y \)[/tex]:
1. For [tex]\( x = -1 \)[/tex]:
[tex]\[ y = 2(-1)^2 - 5(-1) - 12 \][/tex]
[tex]\[ y = 2(1) + 5 - 12 \][/tex]
[tex]\[ y = 2 + 5 - 12 \][/tex]
[tex]\[ y = -5 \][/tex]
2. For [tex]\( x = 1 \)[/tex]:
[tex]\[ y = 2(1)^2 - 5(1) + q \][/tex]
[tex]\[ y = 2(1) - 5(1) - 12 \][/tex]
[tex]\[ y = 2 - 5 - 12 \][/tex]
[tex]\[ y = -15 \][/tex]
3. For [tex]\( x = 2 \)[/tex]:
[tex]\[ y = 2(2)^2 - 5(2) - 12 \][/tex]
[tex]\[ y = 2(4) - 5(2) - 12 \][/tex]
[tex]\[ y = 8 - 10 - 12 \][/tex]
[tex]\[ y = -14 \][/tex]
4. For [tex]\( x = 3 \)[/tex]:
[tex]\[ y = 2(3)^2 - 5(3) - 12 \][/tex]
[tex]\[ y = 2(9) - 5(3) - 12 \][/tex]
[tex]\[ y = 18 - 15 - 12 \][/tex]
[tex]\[ y = -9 \][/tex]
Updating the table:
[tex]\[ \begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline x & -3 & -2 & -1 & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline y & 21 & 6 & -5 & -12 & -15 & -14 & -9 & 0 & 13 \\ \hline \end{array} \][/tex]
b) Graphing the relation
Once you have completed the table, plot the points [tex]\((-3, 21)\)[/tex], [tex]\((-2, 6)\)[/tex], [tex]\((-1, -5)\)[/tex], [tex]\((0, -12)\)[/tex], [tex]\((1, -15)\)[/tex], [tex]\((2, -14)\)[/tex], [tex]\((3, -9)\)[/tex], [tex]\((4, 0)\)[/tex], and [tex]\((5, 13)\)[/tex] on a graph using the given scales.
c) Use the graph to find:
(i) [tex]\( y \)[/tex] when [tex]\( x = 1.8 \)[/tex]:
Reading the value from the graph or using the equation:
[tex]\[ y = 2(1.8)^2 - 5(1.8) - 12 \][/tex]
[tex]\[ y = 2(3.24) - 9 - 12 \][/tex]
[tex]\[ y = 6.48 - 9 - 12 \][/tex]
[tex]\[ y = -14.52 \][/tex]
Hence, [tex]\( y = -14.52 \)[/tex].
(ii) [tex]\( x \)[/tex] when [tex]\( y = -8 \)[/tex]:
Reading the value from the graph, we see that there are no [tex]\( x \)[/tex] values corresponding to [tex]\( y = -8 \)[/tex]. This indicates the curve does not intersect [tex]\( y = -8 \)[/tex] within the given range of [tex]\( x \)[/tex] values.
Thus, [tex]\( x \)[/tex] when [tex]\( y = -8 \)[/tex] has no valid solutions in the specified domain.
