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3 years ago

Which is a possible turning point for the continuous function f(x)?

Mathematics

1 answer:

babymother [125]3 years ago

5 0

Answer:

We are given coordinates of a continuous function f(x)

(–2, 0)

(0, –2)

(2, –1)

(4, 0).

We need to find the possible turning point for the continuous function.

Note: Turning point is a point on the graph where slope of the curve changes from negative to positive or positive to negative.

A turning point is always lowest or highest point of the curve (where bump of the graph seen).

For the given coordinates we can see that (–2, 0) and (4, 0) coordinates are in a same line, that is on the x-axis.

But the coordinate (0, –2) is the lowest point on the graph.

Therefore, (0, –2) is the turning point for the continuous function given.

hoped this was helpful!

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Step-by-step explanation:

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1) $K' = \alpha_{1} \cdot K$

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3) $P' = \alpha_{3} \cdot P$

4) $J' = \alpha_{4} \cdot J$

5) $\alpha_{1} = \alpha_{2} = \alpha_{3} = \alpha_{4}$

If we know that $K = (2, 0)$ , $K' = (1, 0)$ , $T = (3, 0)$ , $T' = (1.5,0)$ , $P = (1, 5)$ , $P' = (0.5, 2.5)$ , $J = (0, 3)$ and $J' = (0, 1.5)$ , then the coefficients are, respectively:

$\alpha_{1} = \frac{1}{2}$ , $\alpha_{2} = \frac{1}{2}$ , $\alpha_{3} = \frac{1}{2}$ , $\alpha_{4} = \frac{1}{2}$

As $\alpha_{1} = \alpha_{2} = \alpha_{3} = \alpha_{4}$ , we conclude that the dilation scale factor applied in the preimage is equal to $\frac{1}{2}$ .

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