Answer:
(a) The table
[tex]\begin{array}{cc}x & {f(x)} & {0} & {0} & {1} & {7}& {2} & {12} & {3} & {15} & {4} & {16} & {5} &{15} & {6} & {12} & {7} &{7} & {8} & {0}\ \end{array}[/tex]
(b) See attachment for graph
(c) He swapped the values of the x and y coordinates for one another
Step-by-step explanation:
Given
See attachment for complete question
Solving (a): Complete the table
[tex]f(x) = -x^2 + 8x[/tex]
[tex]x = 0[/tex]: [tex]f(0) = -0^2 + 8*0 = 0 + 0 = 0[/tex]
[tex]x = 1[/tex]: [tex]f(1) = -1^2 + 8*1 = -1 + 9 = 7[/tex]
[tex]x = 2[/tex]: [tex]f(2) = -2^2 + 8 *2 = -4 + 16 = 12[/tex]
[tex]x = 3[/tex]: [tex]f(3) = -3^2 + 8 *3 = -9 + 24 = 15[/tex]
[tex]x = 4[/tex]: [tex]f(4) = -4^2 + 8 *4 = -16 + 32 = 16[/tex]
[tex]x = 5[/tex]: [tex]f(5) = -5^2 + 8 *5 = -25 + 40 = 15[/tex]
[tex]x = 6[/tex]: [tex]f(6) = -6^2 + 8 *6 = -36 + 48 = 12[/tex]
[tex]x= 7[/tex]: [tex]f(7) = -7^2 + 8 *7 = -49 + 56 = 7[/tex]
[tex]x = 8[/tex]: [tex]f(8) = -8^2 + 8 *8 = -64 + 64 = 0[/tex]
So, the complete table is:
[tex]\begin{array}{cc}x & {f(x)} & {0} & {0} & {1} & {7}& {2} & {12} & {3} & {15} & {4} & {16} & {5} &{15} & {6} & {12} & {7} &{7} & {8} & {0}\ \end{array}[/tex]
Solving (b): Plot the graph
See attachment 2 for graph
Solving (c): Why he was marked as incorrect
By comparing the new attached graph (in b) and the graph of the original question, one will notice that the values of the x and y coordinates are switched.
This is why he was marked as incorrect.
