Question 1204096: how do you turn the vertex form; y=4(3x-1)^2+8 back into standard form?
Answer by math_tutor2020(3816)   (Show Source):
You can put this solution on YOUR website!
Answer: y = 36x^2-24x+12
Work Shown
y = 4(3x-1)^2+8
y = 4(3x-1)(3x-1)+8
y = 4(9x^2-6x+1)+8 ....... use the FOIL rule or box method
y = 36x^2-24x+4+8 ....... distribute
y = 36x^2-24x+12
We have 4(3x-1)^2+8 expand and simplify to 36x^2-24x+12
Therefore, 4(3x-1)^2+8 = 36x^2-24x+12 is an identity.
It is a true equation for all real numbers x.
Ways to verify:- Graph y = 4(3x-1)^2+8 and y = 36x^2-24x+12 to notice they produce the exact same parabola (i.e. they pass through the exact same set of points). Desmos and GeoGebra are two graphing options I recommend. Graphing calculators such as a TI83 or TI84 work as well.
- Make a table of values comparing each quadratic. The tables should be identical when using the same input x values.
- Use a computer algebra system (CAS) to verify that 4(3x-1)^2+8 turns into 36x^2-24x+12. GeoGebra is one example of a CAS. WolframAlpha is another.
Extra side notes:- Standard form is y = ax^2 + bx + c
- Technically the given equation y = 4(3x-1)^2+8 is NOT in vertex form because vertex form has the template y = a(x-h)^2+k
- The vertex form for this particular equation is y = 36(x - 1/3)^2 + 8
- The vertex is located at (1/3, 8)
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