My equation is : x=1/4(x^2-2x+5). I need to get it into standard form so I can finish solving the parabola...?

Question

I'm just having a problem getting it into standard form...probably because I've spent so long away from this kind of math. Help?

Answer 1

Did you mean y=... rather than x=..?

If you meant y=..., to get it in standard form, complete the square.

Looking inside the parens, the coefficient of x is -2. Divide by 2: -1
Square: 1

Now add and subtract 1 inside the parens:

y=1/4[(x^2-2x+1)-1+5]

Now, the first three terms inside the parens are a perfect square:

x^2-2x+1=(x-1)^2.

Thus,

y=1/4[(x^2-2x+1)-1+5]
y=1/4[(x-1)^2+4]
y=1/4(x-1)^2+1

Thus, the graph of the parabola goes up (because the lead coefficient is positive) and has vertex and (1,1).

Since it's vertex is above the x-axis and it goes up, there are no x-intercepts. That is, y=0 does not have any real solutions.

Note that in general, given any quadratic, you can always complete the square.

If y=ax^2+bx+c, with a not equal to zero, first divide by a:
y/a=x^2+b/a x+c/a.

The coefficient of x is b/a. Divide by 2, b/( 2 a) and square this: b^2/(4a^2).

Now add and subtract this number on the right side of the equation:

y/a=x^2+b/a x+b^2/(4a^2)+c/a-b^2/(4a^2)

The first three terms on the right side of the equation are a perfect square

y/a=(x+b/(2a) )^2+(4ac-b^2)/(4a^2)

Multiply through by a:

y=a (x+b/(2a) )^2+(4ac-b^2)/(4a)

This shows you that the vertex of the parabola is (-b/2a,(4ac-b^2)/(4a)). The parabola goes up if a is positive and down if a is negative.

Note that setting y=0 and solving for x results in the quadratic formula.

Answer 2

The equation given is not the equation of a parabola. It is simply a quadratic equation in x. If the equation were y=1/4(x^2-2x+5) then it would the equation of a parabola which could be rewritten as y=1/4(x-1)^2 +1.

Answer 3

you want to isolate the x so as which you have x= some thing. as a fashion to try this, you upload 5 to the two facets of the equation (because of the fact 5-5=0) with the aid of doing which you get 2x-5+5=12+5 Now you have 2x=17 Now, you only want the x term, so which you divide the two facets of the equation with the aid of 2 (because of the fact 2/2=a million) with the aid of doing which you get x=17/2=8.5

Answer 4

I would solve it in the following way

x = 1/4(x^2 - 2x + 5)
multiply both sides by 4(x^2 - 2x +5)
4x(x^2 - 2x +5) = 1
4x^3 - 8x^2 + 20x = 1
4x^3 - 8x^2 + 20x -1 =0

then factorise to get your result

Answer 5

i don't know hat you mean by standard form but,
if x=1/4(x^2-2x+5)
4x=x^2-2x+5
0=x^2-6x+5
0=(x-5)(x-1)
so x= 5 or 1

Answer 6

Why don't you spend your time doing something a little more practical? I mean seriously, when in the real world would you EVER need to know the answer to a problem like this?

It's just a waste of time.

Answer 7

(x-5)(x-1)=0

Answer 8

Huh?