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3 years ago

Logbase3 (x+2)= log base3 (2x^2-1) steps 1 through 6

1 answer:

5 0

$log_3(x+2)=log_3(2x^2-1)\\3^{log_3(x+2)}=3^{log_3(2x^2-1)}\\x+2=2x^2-1\\-x -2 -x -2\\0=2x^2-x-3\\2*-3=-6\\Test factors -3 and 2:\\-3*2=-6, -3+2=-1\\(x-3)(x+2)=0\\Use slide and divide with 2:\\(x-3/2)(x+2/2)=0\\=>\\(2x-3)(x+1)=0\\Set both equal to 0:\\x+1=0\\x=-1\\2x-3=0\\2x=3\\/2 /2\\x=3/2\\$

Answer:

x=-1, 3/2

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For a and b, write an equation in slope-intercept form that meets the given criteria

MariettaO [177]

Answer:

A) y = -2x

B) y = x - 3

Step-by-step explanation:

A) you need to have a negative 'x' coefficient and no y-intercept

B) you need to have a positive 'x' coefficient and a negative y-intercept

8 0

2 years ago

1. Write the equation of the piece-wise function graphed below.

DaniilM [7]

Problem 4

<h3>Answer:</h3>

$f(x) = \begin{cases}2x+4 \text{ if } x < 1\\x-3 \text{ if } x \ge 1\end{cases}$

------------------

Work Shown:

The left line goes through (-2,0) and (0,4)

The slope of this line is

m = (y2-y1)/(x2-x1)

m = (4-0)/(0-(-2))

m = (4-0)/(0+2)

m = 4/2

m = 2

The y intercept is b = 4

Since m = 2 and b = 4, this means y = mx+b turns into y = 2x+4. This portion is only done when x < 1. Note the open circle at the endpoint of this portion. So we do not include x = 1 as part of this piece.

---

The line on the right side goes through (1,-2) and (2,-1)

Slope

m = (y2-y1)/(x2-x1)

m = (-1-(-2))/(2-1)

m = (-1+2)/(2-1)

m = 1/1

m = 1

The y intercept is b = -3. You can see this if you extend the line until it crosses the y axis.

Alternatively, plug in (x,y) = (1,-2) and m = 1 into y = mx+b to find that b = -3

So y = mx+b turns into y = 1x+(-3) or just y = x-3

We combine both parts to end up with $f(x) = \begin{cases}2x+4 \text{ if } x < 1\\x-3 \text{ if } x \ge 1\end{cases}$

This is only graphed when $x \ge 1$ (note the closed or filled in circle for the endpoint of this portion).

===================================================

Problem 5

Answer:

<h3> $f(x) = \frac{1}{2}|x+3|$ is the absolute value function</h3><h3>while this is the piecewise function</h3>

$f(x) = \begin{cases}-\frac{1}{2}(x+3) \text{ if } x < -3\\\frac{1}{2}(x+3) \text{ if } x \ge -3\end{cases}\\$

------------------

Work Shown:

y = |x| .... parent function

y = |x+3| ... shift 3 units to the left

y = (1/2)*|x+3| .... vertically compress by factor of 1/2

f(x) = (1/2)*|x+3|

------

Break that down into a piecewise function

when x < -3, then y = -(1/2)(x+3)

when $x \ge -3$ , then y = (1/2)(x+3)

I'm using the rule that y = |x| turns into y = -x when x < 0 and y = x when $x \ge 0$

So that is how we get $f(x) = \begin{cases}-\frac{1}{2}(x+3) \text{ if } x < -3\\\frac{1}{2}(x+3) \text{ if } x \ge -3\end{cases}\\$ as the piecewise function.

8 0

3 years ago

Read 2 more answers

What is the answer to 2x^2+8x-64=0 ?

klio [65]

Answer:

Move all terms to the left side and set equal to zero. Then set each factor equal to zero.

x=4, -8

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers