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Alchen [17]
3 years ago
12

The vertex of a parabola is (-1.5, -12.5), and its y-intercept is (0, -8)

Mathematics
1 answer:
Olenka [21]3 years ago
4 0
Best to find the equation of the parabola first:

Its equation is y+12.5 = (x+1.5)^2, or y = -12.5 + (x+1.5)^2

Check the first possible x-intercept:  let x = -2.  Does y come out to 0?

y = -12.5 + (-0.5)^2  =  -12.5 + 0.25.  This is not equal to 0, so (-2,0) is not an x-intercept of this parabola.  Continue checking each possible x-intercept until you find the right one (or ones).
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Answer:

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Step-by-step explanation:

One way: Look at graph

Look at for a horizontal or vertical line that both lines don't touch.

Second Way:

When you look for vertical asymptotes, you set the denominator equal to zero and solve.

For example, the second question has a denominator of x-3.

x-3=0

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When you look for horizontal asymptote, you do long division.

For example, the first question f(x)=\frac{2x}{x-3}

The answer will be 2. 2 is the quotient. You only want quotient not remainder.

y=2.

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Step-by-step explanation:

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Read 2 more answers
Step by step workout of ;<br> log x^3+log 5x=5 log 2-log 2/5
Charra [1.4K]

Answer:

x = 2

Step-by-step explanation:

log  \: {x}^{3}  + log \: 5x = 5log \: 2 - log \:  \frac{2}{5}  \\  \\  \therefore \: log( {x}^{3}  \times 5x) = log  {2}^{5} - log \:  \frac{2}{5}  \\  \\ \therefore \: log( 5{x}^{4}) = log 32 - log \:  \frac{2}{5} \\  \\ \therefore \: log( 5{x}^{4}) = log  \bigg(32  \times  \frac{5}{2}  \bigg) \\  \\ \therefore \: log( 5{x}^{4}) = log  \bigg(16  \times  5) \\  \\ \therefore \: 5{x}^{4}=  16  \times  5 \\  \\ \therefore \: {x}^{4}=   {2}^{4}   \\  \\  \huge \purple{ \boxed{\therefore \: {x} = 2}}

4 0
3 years ago
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