Directions: Graph each equation by using the y-intercept and slope( make sure too show your work)

Mathematics

1 answer:

vova2212 [387]3 years ago

4 0

Given:

The equation of a linear function is:

$y=-\dfrac{2}{3}x-1$

To find:

The y-intercept and slope of the given equation, then draw the graph of the equation.

Solution:

Slope intercept form of a line is:

$y=mx+b$ ...(i)

Where, m is the slope and b is the y-intercept

We have,

$y=-\dfrac{2}{3}x-1$ ...(ii)

On comparing (i) and (ii), we get

$m=-\dfrac{2}{3}$

$b=-1$

Therefore, the y-intercept is -1 and the slope of the line is $-\dfrac{2}{3}$ .

It means the graph intersect the y-axis at point (0,-1) and having slope $-\dfrac{2}{3}$ . So, the graph of the equation is shown below.

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WILL MARK BRAINLIEST what is the domain of (f/g) (x)?

tatyana61 [14]

Given that $f(x) = \sqrt{7-x}$ and $g(x) = \sqrt{x + 2}$ , we can say the following:

$\Bigg(\dfrac{f}{g}\Bigg)(x) = \dfrac{f (x)}{g(x)} = \dfrac{\sqrt{7 - x}}{\sqrt{x+2}}$

Now, remember what happens if we have a negative square root: it becomes an imaginary number. We don't want this, so we want to make sure whatever is under a square root is greater than 0 (given we are talking about real numbers only).

Thus, let's set what is under both square roots to be greater than 0:

$\sqrt{7 - x} \Rightarrow 7 - x \geq 0 \Rightarrow x \leq 7$

$\sqrt{x + 2} \Rightarrow x + 2 \geq 0 \Rightarrow x \geq -2$

Since both of the square roots are in the same function, we want to take the union of the domains of the individual square roots to find the domain of the overall function.

$x \leq 7 \,\,\cup x \geq -2 = \boxed{-2 \leq x \leq 7}$

Now, let's look back at the function entirely, which is:

$\Bigg( \dfrac{f}{g} \Bigg)(x) = \dfrac{\sqrt{7 - x}}{\sqrt{x+2}}$

Since $\sqrt{x + 2}$ is on the bottom of the fraction, we must say that $\sqrt{x + 2} \neq 0$ , since the denominator can't equal 0. Thus, we must exclude $\sqrt{x + 2} = 0 \Rightarrow x + 2 = 0 \Rightarrow x = -2$ from the domain.