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Hoochie [10]
3 years ago
8

Was i right on the first 2? Please correct them if i'm wrong

Mathematics
1 answer:
Blababa [14]3 years ago
7 0
All correct!!! Good job!!! ^^
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3 years ago
100 POINTS!!!!!!!!!!!!!!!!!!!!!!!!!!! PLEASE SHOW STEPS TO ALL PARTS
goldfiish [28.3K]

Answer:

Step-by-step explanation:

<u>Part A: the x-intercepts are -3/2 and 5/2 </u>

  • 4x^{2} - 7x - 15 -----> (2x + 3) (2x - 5)

2x + 3 = 0 --------> 2x = -3 ----------> x = \frac{-3}{2}

2x - 5 = 0 ---------> 2x = 5 -----------> x = \frac{5}{2}

<u>Part B: the parabola is a minimum, the vertex is (7/8,-289/16) </u>

f(x)=ax^2+bx+c

if a>0, then the parabola opens up and the vertex is a minimum

a<0 then the parabola opens down and the vertex is a max

  • a = 4, it is a minimum

the x value of  the vertex in f(x)=ax^2+bx+c= is -b/(2a)

the y value of the vertex is f(-b/(2a))  

f(x) = 4x^{2} - 7x - 15

a = 4

b = -7

-b/2a=-(7)/(2*4)=7/8

f(7/8) = 4(7/8)^{2} - 7(7/8) - 15

f(7/8) = 49/16 - 49/8 - 15

f(7/8) = -289/16

  • the vertex is (7/8,-289/16)

<u>Part C: </u>

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3 years ago
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