Question

State the vertex of the parabola. y=2x^2-12x+5

Answer 1

2 options

1. complete the square in x.

2. find the line of symmetry and substitute that value into the equation. The vertex is on the line of symmetry....

I would choose option 2.

for a parabola

\[y = ax^2 + bx + c\]

the line of symmetry is \[x = \frac{-b}{2 \times a}\]

in your question a = 2 and b = -12

substitute the values to find the line of symmetry.

When you get it, substitute the value into the original equation.

the vertex will be (x, f(x))

hope it helps

Answer 2

(3,-13)
Refer to the attachment from Mathematica 9.

Answer 3

i'd suggest completing the square as it is a fundamental approach where you drive the process and get to understand what is really going on. moreover, it is also the derivation of the - b +/- sqrt(b^2 - 4ac) etc etc stuff that is so useful.

here goes:

y = 2x^2 - 12x + 5
y = 2{x^2 - 6x + 5/2}
now, we know (x-3)^2 = x^2 - 6x + 9
so y = 2 {(x-3)^2 - 9 + 5/2 } = 2 { (x-3)^2 -13/2 } = 2(x-3)^2 - 13

CHECK x = 3, y = -13 in both forms: TICK

a more modularised approach is often recommended online, as follows:

STEP 1: y - 5 = 2x^2 -12x [ie leave only x^2 and x on RHS getting ready for step 3]
STEP 2: (1/2)(y-5) = x^2 - 6x [ie x^2 coefficient set to 1 getting ready for step 3]
STEP 3: (1/2)(y-5) = (x-3)^2 -9 [complete square on LHS!!!!]
STEP 4: y-5 = 2(x-3)^2 - 18 [rest is just working back]
STEP 5: y = 2(x-3)^2 -13

i prefer the first approach, personally, as i believe maths is about understanding and not a memory test in formulae / methods / tricks...

either way, we have restated the parabola around x = 3, and can compare it in our heads to the simplest form, which is y = x^2.

we can then see that it mins or maxes at x = 3, and we know that its min or max is y = -13. furthermore because it is +2 times (x-3)^2 we know that it is a min, as with the simple y = x^2.

Answer 4

wow... interesting approach I would have simply isolated the terms is x then factored and completed the square

\[y = 2(x^2 - 6x) + 5\]

avoids fractions...

and it was an interesting comment about a fundemental approach to solving

given the general quadratic formula

\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

which could be written as

\[x = \frac{-b}{2a} \times (\pm \sqrt{b^2 - 4ac})\]

wow... the 1st part is the line of symmetry... go figure...

and the 2nd part \[\pm \sqrt{b^2 - 4ac}\] is telling you to find 2 values that are equidistant from the line of symmetry... for the zeros of the parabola

isn't that amazing...

for me... keeping it simple... works best...

get the fundementals correct and use them over and over again... its like a builder going to his toolbox...

Answer 5

and oh wow again... I didn't give an answer....

Answer 6

I'll probably have my teamwork score downgraded again for a cheeky response...

Answer 7

lol!!