Question 1204096
<font color=black size=3>
Answer:   <font color=red size=4>y = 36x^2-24x+12</font>


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 <a href = "https://www.algebra.com/algebra/homework/playground/lessons/box-method.lesson">box method</a>
y = 36x^2-24x+4+8 ....... distribute
<font color=red>y = 36x^2-24x+12</font>


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:<ol><li>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.</li><li>Make a table of values comparing each quadratic. The tables should be identical when using the same input x values.</li><li>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.</li></ol>
Extra side notes:<ul><li>Standard form is y = ax^2 + bx + c</li><li>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</li><li>The vertex form for this particular equation is y = 36(x - 1/3)^2 + 8</li><li>The vertex is located at (1/3, 8)</li></ul>
</font>