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Advocard [28]
3 years ago
13

What is the vertex form, f(x) = a(x - h)2-k, for a parabola that passes through the point (2,8) and has (1,0) as its

Mathematics
1 answer:
Novay_Z [31]3 years ago
3 0

Answer:

Step-by-step explanation:

hello :

f(x) = a(x - h)²-k   ....vertex form  when the vertex is : (h,k)

h=1  and  k=0   so : f(x) = a(x-1)²

a parabola that passes through the point (2,8) : f(2)=8

a(2-1)² =8

a= 8

f(x) = 8(x-1)²=  8(x²-2x+1)

f(x) 8x²-16x+8   .....standard form

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Discuss the continuity of the function on the closed interval.Function Intervalf(x) = 9 − x, x ≤ 09 + 12x, x > 0 [−4, 5]The f
quester [9]

Answer:

It is continuous since \lim_{x\to 0^{-}} = f(0) = \lim_{x \to 0^{+} f(x)

Step-by-step explanation:

We are given that the function is defined as follows f(x) = 9-x, x\leq 0 and f(x) = 9+12x, x>0 and we want to check the continuity in the interval [-4,5]. Note that this a piecewise function whose only critical point (that might be a candidate of a discontinuity)  x=0 since at this point is where the function "changes" of definition. Note that 9-x and 9+12x are polynomials that are continous over all \mathbb{R}. So F is continous in the intervals [-4,0) and (0,5]. To check if f(x) is continuous at 0, we must check that

\lim_{x\to 0^{-}} = f(0) = \lim_{x \to 0^{+} f(x) (this is the definition of continuity at x=0)

Note that if x=0, then f(x) = 9-x. So, f(0)=9. On the same time, note that

\lim_{x\to 0^{-}} f(x) = \lim_{x\to 0^{-}} 9-x = 9. This result is because the function 9-x is continous at x=0, so the left-hand limit is equal to the value of the function at 0.

Note that when x>0, we have that f(x) = 9+12x. In this case, we have that

\lim_{x\to 0^{+}} f(x) = \lim_{x\to 0^{+}} 9+12x = 9. As before, this result is because the function 9+12x is continous at x=0, so the right-hand limit is equal to the value of the function at 0.

Thus, \lim_{x\to 0^{-}} = f(0) = \lim_{x \to 0^{+} f(x)=9, so by definition, f is continuous at x=0, hence continuous over the interval [-4,5].

5 0
4 years ago
Read 2 more answers
22.  What is an equation of the parabola with vertex at the origin and focus (−4, 0)?
miv72 [106K]
The correct answer for the question shown above is the second option (Option b), and the option cwhich is:

 b) x = -1/16y^2
 c) y=-1/16x^2

 
The explanation is shown below:

 
The focus of a parabola is (h,k+1/4a), then, you have:

 h,k= (0,0)
 1/4a=1/4(-1/16)

 Then, you obtain:

 1/4a=-4

 Therefore, the focus of the equation of the parabola x = -1/16y^2 and the parabola y=-1/16x^2 <span>is F(0-4)
</span>
 So, as you can see, the correct answer is the options mentioned before, the option b and the option c.
 
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3 years ago
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Answer:

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Step-by-step explanation:

<h3>hope it help you !!!:)</h3>
8 0
3 years ago
Whats the table function for y=2x-5
algol13

y = 2x - 5

<u>    x     |     y     </u>

<u>   -1     | 2(-1) - 5 = -7</u>

<u>    0    |  2(0) - 5 = -5</u>

<u>     1    |  2(1) - 5 = -3</u>

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