Find focus of parabola

[frankinaround]

Sorry posting from phone with no symbols. Find the vertex and focus of the parabola. Y=2x squared minus 4x plus 1 i can find the vertex on this but cant figure out the focus. Ive been trying 4 hours. I can figure out the focus on easyer problems though like y=5xsquared focus is 1/20 but yeah dont get it please help. Also the book says the answer is x=2, y=negative2 and 3/4 but dont see how they got that answer.

 

[galactus]

By completing the square with \(\displaystyle y=2x^{2}-4x+1\), we get

\(\displaystyle y=2(x-1)^{2}-1\)

Clearly, the vertex has cooridnates (1,-1)

We have the form \(\displaystyle (x-h)^{2}=4p(y-k)\)

\(\displaystyle (x-1)^{2}=\frac{1}{2}(y+1)\)

\(\displaystyle 4p=\frac{1}{2}\)

\(\displaystyle p=\frac{1}{8}\)

As you know, p is the distance from the vertex to the focus.

Thus, the focus is at \(\displaystyle (1,\frac{-7}{8})\)

If the book says the focus is at x=2, it is in error. Perhaps that is why you have been stuck.

 

[mmm4444bot]

the book says the answer is x=2, y=negative2 and 3/4 but dont see how they got that answer. I do.

Whoever worked this exercise for the text's author copied (or entered) the equation wrongly.

(2, -11/4) is the focus point for the parabola defined by y = x^2 - 4x + 1

 

[frankinaround]

so for x^2-4x+1 the vertext is from y+1=(x-2)^2 so - (2,-1)
then theres nothing outside the x-2, so that means 4p=1/1 ? (because theres no constant b4 y or x, am I wrong? not sure.)
then 1/1*1/4 is p=1/4
then I have to add the distance... so if (2,-1) is the vertex the focus will be (2,-1+1/4) ?

 

[mmm4444bot]

so for x^2-4x+1 the [vertex] is from y+1=(x-2)^2 so - (2,-1) No.

Completing the Square gives: y = (x - 2)^2 - 3

Vertex: (2, -3)



theres nothing outside the x-2, so that means 4p=1/1 ? (because theres no constant b4 y or x, am I wrong? not sure.)

(Please do not type text-speak at this site.)

Yes, 4p = 1

When we see a variable without a coefficient, it does not mean "theres no constant". We understand the constant to be 1, like so:

1y + 3 = 1x^2 - 4x

It works the same way with expressions:

(x - 2)^2 means 1*(x - 2)^2



p=1/4

then I have to add the distance ... so if (2, -3) is the vertex the focus will be (2, -3 + 1/4) ? Yes, with corrections shown in red.

Of course, you will simplify the expression -3 + 1/4, yes?

(2, -11/4)