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uysha [10]
3 years ago
7

Determine the verticle asymptote and horizontal asymptote of x^2-25/2x^+13x+15?

Mathematics
1 answer:
In-s [12.5K]3 years ago
5 0

Answer:

what are you doing here honey and what is this gonna be about me in a hurry for the first two

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Help Me with the second one plz!
Hoochie [10]

Answer:

0.003

Step-by-step explanation:

multiply by 12

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Write the equation of the function whose graph is shown. y = (x + )2 +
Deffense [45]
Quadratic form is given in the form
y= ax^{2}+bx+c, where a, b, and c are constants

We substitute the value (8, 12) and (5,3) in turn

12= a(8)^{2}+b(8)+c
12=64a+8b+c ... Equation 1

3=a(5)^{2}+b(5)+c
3=25a+5b+c ... Equation 2

Equation 1 - Equation 2 gives
39a+3b=9, we call this Equation 3

We also know that (5, 3) is the turning point of the curve, where the value of x is given by x=- \frac{b}{2a}
Substitute x=5
5=- \frac{b}{2a}
10a=-b
b=-10a ... Equation 4

Substitute Equation 4 into Equation 3
39a+3b=9
39a+3(-10a)=9
39a-30a=9
9a=9
a=1

Now we need to work out the value of b and c

Substitute a=1 back into equation 4
-10a=b
-10(1)=b
b=-10

Substitute a=1 and b=-10 into either equation 1 or equation 2

64a+8b+c=12
64(1)+8(-10)+c=12
64-80+c=12
c=12+80-64
c=28

Hence the equation of the quadratic curve is
y= x^{2} 10x+28


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3 years ago
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Here is the work to go with the answer it is hard to completely type out hope it helped!

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