JacksonJRB
  • JacksonJRB
Algebraically determine the domain and range: y=x^2-8x+7
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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JacksonJRB
  • JacksonJRB
i already know the domain just need help with the range
jim_thompson5910
  • jim_thompson5910
have you learned about completing the square?
JacksonJRB
  • JacksonJRB
Yes

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More answers

anonymous
  • anonymous
wait you got x you said right?
JacksonJRB
  • JacksonJRB
\[x=y^2-8y+7\]
JacksonJRB
  • JacksonJRB
And yes
jim_thompson5910
  • 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
  • JacksonJRB
I was taught to substitute the x for the y and vice vera
anonymous
  • anonymous
@jim_thompson5910 can't he use substitution?
jim_thompson5910
  • jim_thompson5910
@JacksonJRB you're thinking of the inverse
JacksonJRB
  • JacksonJRB
Ah
jim_thompson5910
  • jim_thompson5910
first compute h = -b/(2a) in this case, a = 1, b = -8
freckles
  • 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
  • jim_thompson5910
once you know the value of h, plug it into y=x^2-8x+7 to find k
SolomonZelman
  • 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
  • jim_thompson5910
`can't he use substitution?` I'm not sure what you mean @GTA_Hunter35
anonymous
  • anonymous
he said he has x
jim_thompson5910
  • jim_thompson5910
He said he had the domain. Not just a single value of x.
anonymous
  • anonymous
nvm it won't work
JacksonJRB
  • JacksonJRB
^^
anonymous
  • anonymous
^_^
anonymous
  • anonymous
its your call @jim_thompson5910
JacksonJRB
  • JacksonJRB
I'm still very confused...
jim_thompson5910
  • jim_thompson5910
were you able to compute h = -b/(2a) ?
freckles
  • freckles
do you want to complete the square or find the zeros or use the vertex formula?
JacksonJRB
  • JacksonJRB
Yes @jim_thompson5910
jim_thompson5910
  • jim_thompson5910
what value did you get for h
anonymous
  • anonymous
wait wats the vertex formula?
JacksonJRB
  • JacksonJRB
And either way @freckles
JacksonJRB
  • JacksonJRB
h=4
freckles
  • freckles
well if you find the average of the zeros you will find the x-coordinate of the vertex
jim_thompson5910
  • jim_thompson5910
plug x = 4 into y=x^2-8x+7 and you get y = ??
freckles
  • freckles
nevermind you guys got the x-coordinate of the vertex
JacksonJRB
  • JacksonJRB
-9
jim_thompson5910
  • 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
  • JacksonJRB
Ah. That makes sense now. Thank you so much!
jim_thompson5910
  • jim_thompson5910
no problem
freckles
  • 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
  • anonymous
well ur right too but u used the function way
freckles
  • freckles
used the function way?
anonymous
  • anonymous
u used f of x
anonymous
  • anonymous
dat was the only difference
freckles
  • freckles
k

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