Write the given quadratic function y=ax²+bx+c to its equivalent standard form y=a(x-h)²+k. 1.) y=x²-x+13/4 2.) y=1/2x²-3x+3 3.) y=-2x²+12x-17 4.) y=x²-4x+1 5.) y=2x²-4x+4 what mathematics concept did you use in doing the transformation ? explain how quadratic function in the form y=ax²+bx+c can transformed into the form y=a(x-h)²+k. PLEASE I NEED HELP !
You would need to use Completing the Square Do you want an example?
yes..
please .. help me ... @ganeshie8 @Ahsome
Ok
Let's do this equation: \[ y=x^2-x+\frac{13}{4}\]
Have you heard of completing the square before?
sorry i haven't ..
Thats gonna be hard
Have you done this before: \[(a+b)^2=a^2+2ab+b^2\]\[(a-b)^2=a^2-2ab+b^2\]
yeah ..
Ok. Completing the square is used to make any equation into that form, so it is easier to simplify.
Equation: \[y=x^2-x+\frac{13}{4}\] Now, to complete the square, we need to make a number. We get this number by doing: \[(\frac{b}{2})^2\]
What is the \(b\) value in this equation?
it's 1.
no, its -1. Can you see that?
oh sorry .. then ?
Ok. Sub that value to the equation: \[(\frac{b}{2})^2\] \[(\frac{-1}{2})^2\] \[0.25\]
Add that number to the equation after the \(x\) \[y=x^2−x+0.25+\frac{13}{4}\] The issue here is that we can't simply just add a number and not change the value. So we need to subtract the same vaule to even it out. I will do that into the end of the equation: \[y=x^2−x+0.25+\frac{13}{4}-0.25\]
Does that make sense?
after that ?
Now. can you see this section: \[x^2-x+0.25\] That part resembles this section: \[a^2-2ab+b^2\] Which we know can simplify to: \[(a-b)^2\] Therefore, we can simplify the equation: \[y=x^2-x+0.25+\frac{13}{4}−0.25\] To: \[y=(x-0.5)^2+\frac{13}{4}−0.25\] Simplify the rest \[y=(x-0.5)^2+3\] Tada. we have now transformed it into standard form :D
thanks @Ahsome . i'll just solve the remaining .. :D thanks for the help ..
No problem @heizl46. If you didn't understand anything, just say so!
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