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Mathematics 13 Online
OpenStudy (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)

OpenStudy (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.

OpenStudy (egracer):

Would it be B?

OpenStudy (anonymous):

Don't think so.

OpenStudy (egracer):

Sorry trying to figure this out haha math isn't my best subject so its a little harder let me try to recalculate

OpenStudy (anonymous):

No problem. Take your time. What do you get for the x-coordinate? (-b/2a)

OpenStudy (egracer):

would the x coordinate be -39/2a?

OpenStudy (anonymous):

Not quite. From looking at the given quadratic, what is the value of a and the value of b?

OpenStudy (anonymous):

a multiplies x^2 and b multiplies the x

OpenStudy (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

OpenStudy (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?

OpenStudy (egracer):

a is 5 and b is -30??

OpenStudy (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?

OpenStudy (anonymous):

Sorry. It's\[x=-\frac{ b }{ 2a }\]

OpenStudy (anonymous):

So you're calculating\[x=-\frac{ b }{ 2a } = -\frac{ -30 }{ 2\left( 5 \right) } = ?\]

OpenStudy (egracer):

-75?

OpenStudy (anonymous):

Nope. What is 2 x 5 ?

OpenStudy (egracer):

10

OpenStudy (anonymous):

Good. That's the denominator. Now, what's -30/10 ?

OpenStudy (egracer):

-3

OpenStudy (anonymous):

Good. Finally, what's -(-3) ?

OpenStudy (egracer):

3

OpenStudy (anonymous):

Excellent. So the x-coordinate of the vertex is 3.

OpenStudy (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?

OpenStudy (egracer):

-18?

OpenStudy (anonymous):

Nope. What is 3^2 ?

OpenStudy (egracer):

9

OpenStudy (anonymous):

Great. Now what is 5 x 3^2 ?

OpenStudy (egracer):

45

OpenStudy (anonymous):

Good. Now what is 30 x 3 ?

OpenStudy (egracer):

90

OpenStudy (anonymous):

Great. Now to put it all together\[f \left( 3 \right) = 5\left( 3 \right)^{2} - 30\left( 3 \right) + 6 = 45-90+6 = ?\]

OpenStudy (egracer):

-39

OpenStudy (anonymous):

Good. Now you have the x- and y-coordinates of the vertex. Where is the vertex?

OpenStudy (egracer):

3,-39 is my answer?

OpenStudy (anonymous):

That's it. Well done.

OpenStudy (egracer):

thank you so much for being patient with me! and helping me understand!

OpenStudy (anonymous):

You're welcome. Good luck with your studies.

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