Algebraically determine the domain and range:
y=x^2-8x+7

- JacksonJRB

Algebraically determine the domain and range:
y=x^2-8x+7

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- JacksonJRB

i already know the domain just need help with the range

- jim_thompson5910

have you learned about completing the square?

- JacksonJRB

Yes

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## More answers

- anonymous

wait you got x you said right?

- JacksonJRB

\[x=y^2-8y+7\]

- JacksonJRB

And yes

- jim_thompson5910

complete the square to get the equation into vertex form y = a(x-h)^2 + k
the range will be \(\Large y \ge k\) (replace k with a numeric value though) because 'a' is positive which means the parabola opens upward

- JacksonJRB

I was taught to substitute the x for the y and vice vera

- anonymous

@jim_thompson5910 can't he use substitution?

- jim_thompson5910

@JacksonJRB you're thinking of the inverse

- JacksonJRB

Ah

- jim_thompson5910

first compute h = -b/(2a)
in this case, a = 1, b = -8

- freckles

there might be an easier way
x^2-8x+7 is factorabole
y=x^2-8x+7
you can find the average of the 0's
so the 0's are a and b
the average of a and b is (a+b)/2
so you can find the range by doing:
\[[f(\frac{a+b}{2}),\infty) \text{ since } a=1>0\]

- jim_thompson5910

once you know the value of h, plug it into y=x^2-8x+7 to find k

- SolomonZelman

y-value of the vertex (the k of the vertex) is the minimum range.
(and there is no maximum limit for range)
no range limits, since it is a polynomial

- jim_thompson5910

`can't he use substitution?`
I'm not sure what you mean @GTA_Hunter35

- anonymous

he said he has x

- jim_thompson5910

He said he had the domain. Not just a single value of x.

- anonymous

nvm it won't work

- JacksonJRB

^^

- anonymous

^_^

- anonymous

its your call @jim_thompson5910

- JacksonJRB

I'm still very confused...

- jim_thompson5910

were you able to compute h = -b/(2a) ?

- freckles

do you want to complete the square or find the zeros or use the vertex formula?

- JacksonJRB

Yes @jim_thompson5910

- jim_thompson5910

what value did you get for h

- anonymous

wait wats the vertex formula?

- JacksonJRB

And either way @freckles

- JacksonJRB

h=4

- freckles

well if you find the average of the zeros you will find the x-coordinate of the vertex

- jim_thompson5910

plug x = 4 into y=x^2-8x+7 and you get y = ??

- freckles

nevermind you guys got the x-coordinate of the vertex

- JacksonJRB

-9

- jim_thompson5910

so k = -9 making the range \(\Large y \ge -9\)
we have a parabola that has the lowest point at (4,-9). All other points will have larger y values.

- JacksonJRB

Ah. That makes sense now. Thank you so much!

- jim_thompson5910

no problem

- freckles

\[f(x)=x^2-8x+7 \\ f(x)=(x-7)(x-1) \\ f(x)=0 \text{ when } x=1 \text{ or } x=7 \\ \text{ the average of the zeros is } \frac{7+1}{2}=\frac{8}{2}=4 \\ \text{ so since } a=1>0 \text{ then the range is } [f(4),\infty) \\ f(4)=4^2-8(4)+7=-9 \\ \text{ so the range is } [-9,\infty)\]
just wanted to type what I was going for

- anonymous

well ur right too but u used the function way

- freckles

used the function way?

- anonymous

u used f of x

- anonymous

dat was the only difference

- freckles

k

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