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# a function reflected

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4

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..

Nov 2, 2018

#1
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$$\text{just complete the square}\\ x^2 - 5x = \\ \\ x^2 - 5x + \dfrac{25}{4} - \dfrac{25}{4} = \\ \\ \left(x-\dfrac 5 2\right)^2 - \dfrac{25}{4}$$

.
Nov 2, 2018
#2
+2

Hi Rom,

but why $$25 \over 4$$?

or $$5 \over2$$ in the first place?..I just don't get this..

juriemagic  Nov 2, 2018
edited by juriemagic  Nov 2, 2018
#3
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$$h(x)=x^2-5x\\$$

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?

Nov 2, 2018
#4
+2

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!!

juriemagic  Nov 2, 2018