How do you get the x intercepts?

Mathematics

1 answer:

Marta_Voda [28]3 years ago

8 0

Factor the equation:
(x-6)(x+1)

Check:
-6+1=-5
-6*1=-6

Use the zero product property
x-6=0
x=6

x+1=0
x=-1

Final answer: x=-1, x=6

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20. Sketch the level curve of the function f(x, y) that contains the point (5,0).

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At the given point,

$f(5,0) = \dfrac1{\sqrt{5^2 + 0^2 - 9}} = \dfrac1{\sqrt{16}} = \dfrac14$

Then the level curve is

$\dfrac1{\sqrt{x^2 + y^2 - 9}} = \dfrac14 \\\\ \implies \sqrt{x^2 + y^2 - 9} = 4 \\\\ \implies x^2 + y^2 - 9 = 4^2 \\\\ \implies x^2 + y^2 = 25 = 5^2$

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2 years ago

Which statements are true of the function f(x) = 3(2.5)x? Check all that apply.

disa [49]

This is an exponential function since the x is in the exponent's place instead of in the place of a "regular" variable. The first statement is true.

The initial value of this particular function is 3 (the other number is the multiplier), so choice 2 is NOT true.

The function increases by its multiplier, which is 2.5, so statement 3 is true.

The equation allows us to enter any x value we want to determine the y, so the domain is in fact all real numbers. So, this statement is also true.

If you were to graph this on a calculator, you would see that the range, the "allowed" y values for our function, do not touch or ever drop below the x-axis. That means that the range is all numbers greater than 0. So that statement is false. No matter what value we pick for x, we will NEVER get back a negative y value or that y = 0. For example, if x = 0, y = 3; if x = -5, y = .03; if x = -10, y = .0003; if x = 5, y = 292.97; if x = -100, $y = 4.8208*88888810^{-40}$ . Y will never be equal to 0 or less than 0.