+1124 hello everyone,
The equation of a parabola is given:
\(f(x)=-x^2-x+6\)
If \(h(x)\) is a reflection of \(f(x)\) about the x-axis, determine it's equation if the graph is shifted 3 units to the right. leave your answer in the form: \(a(x+p)+q\)
Okay, the following I understand:
\(- x^2-x+6\), becomes \(x^2+x-6\) (Reflection)
now add the shift: \({(x-3)}^2+(x-3)-6\)
This equates to: \(x^2-5x\)
NOW THIS PART:
\(h(x)=(x-{5 \over2})^2-{25 \over4}\)
Please explain this last part...Thank you all kindly..
+6248 \(\text{just complete the square}\\ x^2 - 5x = \\ \\ x^2 - 5x + \dfrac{25}{4} - \dfrac{25}{4} = \\ \\ \left(x-\dfrac 5 2\right)^2 - \dfrac{25}{4}\)
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+1124 Hi Rom,
but why \(25 \over 4\)?
or \(5 \over2\) in the first place?..I just don't get this..
+118608
\(h(x)=x^2-5x\\ \)
I'll use a different concrete example to start with
\((x-6)^2=x^2-2*6x+36\\ (x-\frac{12}{2})^2=x^2-12x+(\frac{12}{2})^2\\ x^2-12x+(\frac{12}{2})^2=(x-\frac{12}{2})^2\\~\\ \text{So for your question}\\ x^2-5x+(\frac{5}{2})^2=(x-\frac{5}{2})^2\\ \text{which means}\\ x^2-5x+(\frac{5}{2})^2-(\frac{5}{2})^2=(x-\frac{5}{2})^2-(\frac{5}{2})^2\\ x^2-5x=(x-\frac{5}{2})^2-(\frac{5}{2})^2\\ x^2-5x=(x-\frac{5}{2})^2-(\frac{25}{4})\\ \)
So here, both Rom and I have 'completed the square'
Is that any clearer?
+1124 wow..Melody....yes, it's clearer thank you...I have to admit, I'm going to have to sit with this for a little more, but I'm sure the light will come on!!!... 
..Thank you very much...and to you also Rom..Have a great weekend!!
