egracer
  • egracer
The equation f(x) = 5x2 − 30x + 6 represents a parabola. What is the vertex of the parabola? (−5, 281) (5, −19) (−3, 141) (3, −39)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
In the general equation of a parabola\[f \left( x \right) = ax ^{2} + bx + c\]The x-coordinate of the vertex is given by\[x=-\frac{ b }{ 2a }\]Then calculate f(x) using this value of x to give you the y-coordinate of the vertex.
egracer
  • egracer
Would it be B?
anonymous
  • anonymous
Don't think so.

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egracer
  • egracer
Sorry trying to figure this out haha math isn't my best subject so its a little harder let me try to recalculate
anonymous
  • anonymous
No problem. Take your time. What do you get for the x-coordinate? (-b/2a)
egracer
  • egracer
would the x coordinate be -39/2a?
anonymous
  • anonymous
Not quite. From looking at the given quadratic, what is the value of a and the value of b?
anonymous
  • anonymous
a multiplies x^2 and b multiplies the x
egracer
  • egracer
honestly I have no clue I'm so confused on how to do this could you explain like in details like how to solve it
anonymous
  • anonymous
I just did. But let's back up. The given quadratic function is\[f \left( x \right) = 5x ^{2} - 30x + 6\]Compare that with\[f \left( x \right) = ax ^{2} + bx + c\]a is the number that multiplies x^2 b is the number that multiplies x What are they?
egracer
  • egracer
a is 5 and b is -30??
anonymous
  • anonymous
Perfect. Now the x-coordinate is given by\[x = -\frac{ b }{ a }\]Plug in those values of a and b . What do you get?
anonymous
  • anonymous
Sorry. It's\[x=-\frac{ b }{ 2a }\]
anonymous
  • anonymous
So you're calculating\[x=-\frac{ b }{ 2a } = -\frac{ -30 }{ 2\left( 5 \right) } = ?\]
egracer
  • egracer
-75?
anonymous
  • anonymous
Nope. What is 2 x 5 ?
egracer
  • egracer
10
anonymous
  • anonymous
Good. That's the denominator. Now, what's -30/10 ?
egracer
  • egracer
-3
anonymous
  • anonymous
Good. Finally, what's -(-3) ?
egracer
  • egracer
3
anonymous
  • anonymous
Excellent. So the x-coordinate of the vertex is 3.
anonymous
  • anonymous
Now, to get the y-coordinate of the vertex, put his value of x into the original function and calculate f(3)\[f \left( x \right)=5x ^{2}-30x+6\]\[f \left( 3 \right) = 5\left( 3 \right)^{2}-30\left( 3 \right) + 6\]What do you get?
egracer
  • egracer
-18?
anonymous
  • anonymous
Nope. What is 3^2 ?
egracer
  • egracer
9
anonymous
  • anonymous
Great. Now what is 5 x 3^2 ?
egracer
  • egracer
45
anonymous
  • anonymous
Good. Now what is 30 x 3 ?
egracer
  • egracer
90
anonymous
  • anonymous
Great. Now to put it all together\[f \left( 3 \right) = 5\left( 3 \right)^{2} - 30\left( 3 \right) + 6 = 45-90+6 = ?\]
egracer
  • egracer
-39
anonymous
  • anonymous
Good. Now you have the x- and y-coordinates of the vertex. Where is the vertex?
egracer
  • egracer
3,-39 is my answer?
anonymous
  • anonymous
That's it. Well done.
egracer
  • egracer
thank you so much for being patient with me! and helping me understand!
anonymous
  • anonymous
You're welcome. Good luck with your studies.

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