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JacksonJRB
 one year ago
Algebraically determine the domain and range:
y=x^28x+7
JacksonJRB
 one year ago
Algebraically determine the domain and range: y=x^28x+7

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JacksonJRB
 one year ago
Best ResponseYou've already chosen the best response.0i already know the domain just need help with the range

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2have you learned about completing the square?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wait you got x you said right?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2complete the square to get the equation into vertex form y = a(xh)^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
 one year ago
Best ResponseYou've already chosen the best response.0I was taught to substitute the x for the y and vice vera

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@jim_thompson5910 can't he use substitution?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2@JacksonJRB you're thinking of the inverse

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2first compute h = b/(2a) in this case, a = 1, b = 8

freckles
 one year ago
Best ResponseYou've already chosen the best response.2there might be an easier way x^28x+7 is factorabole y=x^28x+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
 one year ago
Best ResponseYou've already chosen the best response.2once you know the value of h, plug it into y=x^28x+7 to find k

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0yvalue 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
 one year ago
Best ResponseYou've already chosen the best response.2`can't he use substitution?` I'm not sure what you mean @GTA_Hunter35

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2He said he had the domain. Not just a single value of x.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0its your call @jim_thompson5910

JacksonJRB
 one year ago
Best ResponseYou've already chosen the best response.0I'm still very confused...

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2were you able to compute h = b/(2a) ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2do you want to complete the square or find the zeros or use the vertex formula?

JacksonJRB
 one year ago
Best ResponseYou've already chosen the best response.0Yes @jim_thompson5910

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2what value did you get for h

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wait wats the vertex formula?

JacksonJRB
 one year ago
Best ResponseYou've already chosen the best response.0And either way @freckles

freckles
 one year ago
Best ResponseYou've already chosen the best response.2well if you find the average of the zeros you will find the xcoordinate of the vertex

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2plug x = 4 into y=x^28x+7 and you get y = ??

freckles
 one year ago
Best ResponseYou've already chosen the best response.2nevermind you guys got the xcoordinate of the vertex

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2so 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
 one year ago
Best ResponseYou've already chosen the best response.0Ah. That makes sense now. Thank you so much!

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[f(x)=x^28x+7 \\ f(x)=(x7)(x1) \\ f(x)=0 \text{ when } x=1 \text{ or } x=7 \\ \text{ the average of the zeros is } \frac{7+1}{2}=\frac{8}{2}=4 \\ \text{ so since } a=1>0 \text{ then the range is } [f(4),\infty) \\ f(4)=4^28(4)+7=9 \\ \text{ so the range is } [9,\infty)\] just wanted to type what I was going for

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well ur right too but u used the function way

freckles
 one year ago
Best ResponseYou've already chosen the best response.2used the function way?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dat was the only difference
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