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Thepotemich [5.8K]

2 years ago

5

HELP ME ASAP WITH 8! PLS!! I beg u..

1 answer:

lakkis [162]2 years ago

7 0

7. x=3 is the midpoint between the roots. The other root is x = 2*3 -(-5) = 11.

8a) f(x) = (x +3)^2 -49. The vertex is (-3, -49). The roots are -10, 4.

8b) y = (x+4)^2 -1. The vertex is (-4, -1). The roots are -5, -3.

8c) f(x) = 2(x +3)^2 -34. The vertex is (-3, -34). The roots are -3±√17.

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SpyIntel [72]

Answer:

The domain of the function is all real values of x, except $x = 4$ and $x = 2$

Step-by-step explanation:

We are given the following function:

$f(x) = \frac{x+1}{x^2-6x+8}$

It's a fraction, so the domain is all the real values except those in which the denominator is 0.

Denominator:

Quadratic equation with $a = 1, b = -6, c = 8$

Using bhaskara, the denominator is 0 for these following values of x:

$\Delta = (-6)^2 - 4(1)(8) = 36-32 = 4$

$x_{1} = \frac{-(-6) + \sqrt{4}}{2} = 4$

$x_{2} = \frac{-(-6) - \sqrt{4}}{2} = 2$

The domain of the function is all real values of x, except $x = 4$ and $x = 2$

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Read 2 more answers