Answer:
k(x) = x^2 and j(x) = -x^2
Step-by-step explanation:
The higher the coefficient, the steeper the graph. So the ones that work have to be less than 2.
It can't be 3x^2 because 3 is greater than 2x^2.
It can't be -3x^2 because -3 is the same as 3 except on the other side of the graph, and 3 is greater than 2.
It can't be -2x^2 because it has equal steepness as 2x^2, except it's on the opposite side of the graph.
It is x^2 because 1 is less than 2 in 2x^2, so it's less steep.
It is -x^2 because -1 is the same as 1 except on the other side of the graph, and 1 is less than 2.
Hope this helps!
A function maps a given input to a single output, increasing the values of
the factors increases the output and steepness of the function.
The functions that have graphs that have graphs that are less steep than f(x) = 2·x² are; k(x) = x², and j(x) = -x²
Reasons:
The steeper the graph, the larger the ratio of the Rise to Run of the graph.
Therefore, increasing the multiple of the input of the graph increases the
Rise to Run ratio and therefore the steepness of the graph as follows;
Between points x = 1, and x = 2, for m(x) = 3·x², we have;
[tex]Slope = \dfrac{m(2) - m(1)}{2 - 1} = \dfrac{3 \times 2^2 -3 \times 1^2 }{2 - 1} = \dfrac{12 - 3}{2 - 1} = 9[/tex]
For k(x) = x², we have;
[tex]Slope = \dfrac{k(2) - k(1)}{2 - 1} = \dfrac{ 2^2 - 1^2 }{2 - 1} = \dfrac{4 - 1}{2 - 1} = 3[/tex]
Therefore;
The graphs of k(x) = x², and j(x) = -x², are less steep than f(x) = 2·x².
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