4 years ago

For a quadratic function ƒ(x)=ax2+bx+c, ƒ(−1)<1, ƒ(1)>−1, and ƒ(3)<−4. What could be the sign of a?

Mathematics

1 answer:

denis23 [38]4 years ago

4 0

Answer:

"a" must be negative

Step-by-step explanation:

The slope between points (-1, 1) and (1, -1) is a minimum of (-1-1)/(1-(-1)) = -1.

The slope between points (1, -1 and (3, -4) is a maximum of (-4-(-1))/(3-1) = -3/2.

Thus, as x increases, the slope is decreasing. In order for that to be the case, the value of <em>a</em> must satisfy a < 0.

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fomenos

Answer: 0.1 or 1/10

Step-by-step explanation:

$\left(\frac{3}{4}-\frac{2}{3}\right)\cdot \:1\frac{1}{5}$

$1\frac{1}{5}=\frac{6}{5}$

$\left(\frac{3}{4}-\frac{2}{3}\right)\cdot \frac{6}{5}$

$\frac{3}{4}-\frac{2}{3}$ $=\frac{9}{12}-\frac{8}{12}$

$=\frac{1}{12}$

$\frac{6}{5}\cdot \frac{1}{12}$

Cross, cancel common factor

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$=\frac{1}{10}$

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yaroslaw [1]

Answer:

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