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17 votes
17 votes
Pierre was plotting the quadratic function f(x) = - x ^ 2 + 8x for an exit ticket His work is shown below .

User Ant P
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1 Answer

25 votes
25 votes

Answer:

(a) The table


\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}

(b) See attachment for graph

(c) He swapped the values of the x and y coordinates for one another

Explanation:

Given

See attachment for complete question

Solving (a): Complete the table


f(x) = -x^2 + 8x


x = 0:
f(0) = -0^2 + 8*0 = 0 + 0 = 0


x = 1:
f(1) = -1^2 + 8*1 = -1 + 9 = 7


x = 2:
f(2) = -2^2 + 8 *2 = -4 + 16 = 12


x = 3:
f(3) = -3^2 + 8 *3 = -9 + 24 = 15


x = 4:
f(4) = -4^2 + 8 *4 = -16 + 32 = 16


x = 5:
f(5) = -5^2 + 8 *5 = -25 + 40 = 15


x = 6:
f(6) = -6^2 + 8 *6 = -36 + 48 = 12


x= 7:
f(7) = -7^2 + 8 *7 = -49 + 56 = 7


x = 8:
f(8) = -8^2 + 8 *8 = -64 + 64 = 0

So, the complete table is:


\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}

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.

Pierre was plotting the quadratic function f(x) = - x ^ 2 + 8x for an exit ticket-example-1
Pierre was plotting the quadratic function f(x) = - x ^ 2 + 8x for an exit ticket-example-2
User Matt Evans
by
3.2k points