Solve the quadratic equation 2x^2-4x+1=0
1 answer:
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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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