Question

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

Answer 1

i already know the domain just need help with the range

Answer 2

have you learned about completing the square?

Answer 3

Yes

Answer 4

wait you got x you said right?

Answer 5

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

Answer 6

And yes

Answer 7

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

Answer 8

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

Answer 9

@jim_thompson5910 can't he use substitution?

Answer 10

@JacksonJRB you're thinking of the inverse

Answer 11

Ah

Answer 12

first compute h = -b/(2a)

in this case, a = 1, b = -8

Answer 13

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:
\[[f(\frac{a+b}{2}),\infty) \text{ since } a=1>0\]

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

he said he has x

Answer 18

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

Answer 19

nvm it won't work

Answer 20

^^

Answer 21

^_^

Answer 22

its your call @jim_thompson5910

Answer 23

I'm still very confused...

Answer 24

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

Answer 25

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

Answer 26

Yes @jim_thompson5910

Answer 27

what value did you get for h

Answer 28

wait wats the vertex formula?

Answer 29

And either way @freckles

Answer 30

h=4

Answer 31

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

Answer 32

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

Answer 33

nevermind you guys got the x-coordinate of the vertex

Answer 34

-9

Answer 35

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.

Answer 36

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

Answer 37

no problem

Answer 38

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

Answer 39

well ur right too but u used the function way

Answer 40

used the function way?

Answer 41

u used f of x

Answer 42

dat was the only difference

Answer 43

k