Find x and y intercepts

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Given:
r(x) = [tex]-4x^{2}[/tex] + 16x - 15

My work:

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How is this

[Math Tutor]

Yes, the second image is correct. X-intercept is at -1.5 and -2.5

[Guest]

Your first image makes no sense at all! How did you go from
[tex]-4x^2+ 16x- 15[/tex] to [tex]-4(x^2- 4+ 4)- 15- (-16)[/tex]?
Yes, -4+ 4= 0 but why "4"? And what happened to the "x" multiplying 16?

It looks like you are trying to "complete the square", which is a good idea, but doing it incorrectly!

To "complete the square" write [tex]-4x^2+ 16x[/tex] as [tex]-4(x^2- 4x)[/tex].
Now, you know (or should know!) that [tex](x- a)^2= x^2- 2ax+ a^2[/tex]. Compare "2ax" with "4x". To be the same a must be 2 so [tex]a^2= 4[/tex]. That is, to "complete the square we must add and subtract 4. That may be what you were doing but you were doing it in the wrong place! You need [tex]-4(x^2- 4x)+ 15= -4(x^2- 4x+ 4- 4)- 15[/tex]. Now leave the "+4" in the parentheses to complete the square but take the "-4" out: [tex]-4(x^2- 4x+ 4)+ (-4)(-4)- 15= -4(x- 2)^2+ 16- 15= -4(x- 2)^2+ 1[/tex].

Now, setting that equal to 0, to find the x-intercepts, we have [tex]-4(x- 2)^2+ 1= 0[/tex]. Add [tex]4(x- 2)^2[/tex] to both sides to get [tex]1= 4(x- 2)^2[tex]. Divide both sides by 4: [tex]\frac{1}{4}= (x- 2)^2[/tex]. Take the square root of both sides: [tex]\pm\frac{1}{2}= x- 2[/tex]. Add 2 to both sides: [tex]2\pm \frac{1}{2}= x[/tex].

Taking the "+", [tex]x= 2+ \frac{1}{2}= \frac{4}{2}+ \frac{1}{2}= \frac{5}{2}[/tex].
Taking the "-", [tex]x= 2- \frac{1}{2}= \frac{4}{2}- \frac{1}{2}= \frac{3}{2}[/tex].