Answer:
C) x/2 +2y = 4
Step-by-step explanation:
Of the equations offered, only equation C has the given point as a possible solution.
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Since the graph is not given, we don't know if the given point is on the graphed equation.
Could someone answer asap? I only need to answer part D help would be amazing!!! You will get Brainliest if answered within 10 m
inutes!!!
1 answer:
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Step-by-step explanation:
The first one (f×g)(x) is asking you what is f of x times g of x. This means you multiply the function. You can multiply by the first term (5x) then the second (3) and combine like terms.
Combine like terms
The second one asks you to divide.
Factor out f(x):
From dividing f(x) /g(x)
To
Cancel like terms since (5x+3) is on top and bottom of fraction.
You get 3x+2
Answer:
Answers are in bold
Step-by-step explanation:
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<u>Part 1A</u>
Use the Zero Product Property to find the x-intercepts:
0=-(x+4)(x-1)
0=(x+4)(x-1)
0=(x+4)
-4=x
x=-4
0=(x-1)
1=x
x=1
So the x-intercepts are x=-4 and x=1 which is also (-4,0) and (1,0)
For the y-intercepts, plug in x=0:
f(x)=-(0+4)(0-1)
f(x)=(-4)(-1)
f(x)=4
So the y-intercept is y=4 which is also (0,4)
<u>Part 1B</u>
In general, the axis of symmetry is equal to x=h in the equation y=(x-h)^2+k
For a parabola, the axis of symmetry is equal to x=-b/2a, so we expand the given equation and then plug in the values of "a" and "b" to get the axis of symmetry:
f(x)=-(x+4)(x-1)
f(x)=-(x^2+3x-4)
f(x)=-x^2-3x+4
So now we know a=-1 and b=-3, therefore we can plug them in:
x=-b/2a
x=-(-3)/2(-1)
x=3/-2
x=-3/2
x=-1.5
So the equation of the axis of symmetry is x=-1.5
<u>Part 1C</u>
We'll use the equation x=-b/2a to get the x-coordinate for the vertex (which we already know is x=-1.5), then plug in that value of x in the original equation to get the y-coordinate for the vertex:
f(x)=-(-1.5+4)(-1.5+1)
f(x)=-(2.5)(-0.5)
f(x)=-(-1.25)
f(x)=1.25 <-- y-coordinate
So the coordinates of the vertex are (-1.5,1.25)
<u>Part 1D</u>
See the image below
I'll let someone else do the last 4 questions as I am unable to answer them.
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