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Write the function f(x)6x^2+7x+2 in vertex form, and identify its vertex

User Jakob Lind
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1 Answer

5 votes

Answer:

Explanation:

Complete the square to do this. This one is a bit of a pain because of the 7 as our linear term, but a little hard work never killed anyone!

The rules for completing the square are very strict and cannot be strayed from. The first rule is that the leading coefficient HAS to be a positive 1; ours is a positive 6 so we have to factor it out. Let's do that first then we can get to rule number 2.

Set the quadratic equal to 0, then move the constant over to the other side. This is not a rule, per se, but I teach it this way to avoid unnecessary confusion in an already confusing process.


6x^2+7x=-2 and to factor out the 6:


6(x^2+(7)/(6)x)=-2

Now for rule #2: take half the linear term, square it, and add it to both sides. Our linear term is 7/6. Half of that is 7/12, and 7/12 squared is 49/144. BUT we cannot forget about that 6 hanging around out front, refusing to be ignored. It is a multiplier, so when we add 49/144 inside the parenthesis on the left, we have to add 6(49/144) to the right, giving us:


6(x^2+(7)/(6)x+(49)/(144))=-2+(294)/(144)

We will rewrite now by simplifying. The left side can be written into a perfect square binomial (which is why we did this in the first place) and the right side can be added. The left side becomes


6(x+(7)/(12))^2

The 7/12 comes from the square root of 49/144, and the x comes from the square root of x-squared. The right side requires a common denominator:


-2+(294)/(144)=-(288)/(144)+(294)/(144)=(6)/(144)=(1)/(24)

So now we have


6(x+(7)/(12))^2=(1)/(24)

Put evertyhing back onto the left side and set it back to equal y and you're done!


y=6(x+(7)/(12))^2-(1)/(24)

Because the vertex form of a quadratic does not allow us to stray from the format "x -", we have to interpret our h coordinate as "x - (-7/12))" which is movement to the left. This gives us a vertex of


(-(7)/(12),-(1)/(24))

User AngeloC
by
6.9k points
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