anonymous
  • anonymous
For y = x2 + 6x − 16, Determine if the parabola opens up or down. State if the vertex will be a maximum or minimum. Find the vertex. Find the x-intercepts. Describe the graph of the equation. Show all work and use complete sentences
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
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Jhannybean
  • Jhannybean
We can find the vertex first \[y=x^2+6x-16\]\[y+16=x^2+6x\] now we'll be completing the square . \[y+16=x^2+\color{red}6x\]\[y+16+\color{red}9=x^2+6x+\color{red}{9}\]\[y+25=(x+3)^2\]\[y=(x+3)^2-25\]\[y=(x-(\color{blue}{-3}))+(\color{blue}{-25})\] Your vertex.
Luigi0210
  • Luigi0210
we know from the previous it ups upward
Luigi0210
  • Luigi0210
*opens

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Luigi0210
  • Luigi0210
x intercept when y equals 0
primeralph
  • primeralph
it's simple. If the co-eff of x^2 is positive, it opens up, if negative, opens down.
Luigi0210
  • Luigi0210
we could use the first derivative test to determine if it's a max or min.. since it opens up it's a min..
Jhannybean
  • Jhannybean
The formula \[\large y=\color{purple}a(x-h)^2+k\] is the formula for a quadratic, and your "a" value tells if it opens up or down, your formula \[\large y=\color{purple}{(1)}(x-(-3))^2-25\] the "a" is positive, so your parabola opens upward.
primeralph
  • primeralph
@Jhannybean good job.
Luigi0210
  • Luigi0210
It's her thing :P
Jhannybean
  • Jhannybean
you can find your x-intercept by setting y=0. \[\large (x+3)^2-25=0\]\[\large (x+3)^2=25\]\[\large \sqrt{(x+3)^2}= \pm \sqrt{25}\]\[\large |x+3|= \pm \sqrt{25}\] solving for x would give you your x-intercepts. \[\large |x+3| = \pm 5\]
Luigi0210
  • Luigi0210
0=(x+8)(x-2)
Jhannybean
  • Jhannybean
When describing the graph of the equation, you could use your min to state where the slope is inc/dec, and the domain/range of the function
Luigi0210
  • Luigi0210
@sumsumjoe we're not doing all this work for you if you're not gonna try and understand..
Luigi0210
  • Luigi0210
Well Jhan won't :P
Jhannybean
  • Jhannybean
^_^
Luigi0210
  • Luigi0210
and as always.. great job :D
Jhannybean
  • Jhannybean
Lol, thanks
Luigi0210
  • Luigi0210
I think he's gone :l

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