I think it’s c let me double check just in case
I'll do the first one to get you started
The equation y = x^2+16x+64 is the same as y = 1x^2+16x+64
Compare that to y = ax^2+bx+c and we see that
a = 1
b = 16
c = 64
Use the values of 'a' and b to get the value of h as shown below
h = -b/(2a)
h = -16/(2*1)
h = -8
This is the x coordinate of the vertex.
Plug this x value into the original equation to find the corresponding y value of the vertex.
y = x^2+16x+64
y = (-8)^2 + 16(-8) + 64
y = 0
Since the y coordinate of the vertex is 0, this means k = 0.
The vertex is (h,k) = (-8, 0)
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So we found that a = 1, h = -8 and k = 0
Therefore,
f(x) = a(x-h)^2 + k
f(x) = 1(x-(-8))^2 + 0
f(x) = (x+8)^2
is the vertex form
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<h3>Final answer to problem 1 is f(x) = (x+8)^2 </h3>
C) y=6x^2 is widest among these.
Answer:
2 was not multiplied by 5
Step-by-step explanation:
2(x + 5) = 2x + 5
The correct expansion form distributive property should be :
2(x + 5) = 2*x + 2*5 = 2x + 10
2(x + 5) = 2x + 10
1/4t + 9 = 3
Subtract 9 from both sides
1/2t= -6
Divide 1/2 on both sides so that the only thing remaining on the left side is the variable t.
Final Answer: t= -12
