Math

QuestionRewrite y=3x26x+5y=3 x^{2}-6 x+5 as a(xh)2+ka(x-h)^{2}+k with real constants a,h,ka, h, k.

Studdy Solution

STEP 1

Assumptions1. The given equation is y=3x6x+5y=3x^-6x+5 . We need to express this equation in the form a(xh)+ka(x-h)^+k, where aa, hh, and kk are real constants.

STEP 2

The equation a(xh)2+ka(x-h)^2+k is the standard form of a parabola. To convert the given equation into this form, we need to complete the square.

STEP 3

First, we factor out the coefficient of x2x^2 from the first two terms of the given equation.
y=3(x22x)+5y=3(x^2-2x)+5

STEP 4

Next, we complete the square inside the parentheses by adding and subtracting the square of half the coefficient of xx inside the parentheses.
y=3[(x22x+1)1]+y=3[(x^2-2x+1)-1]+

STEP 5

implify the equation by combining like terms and simplifying the expression inside the parentheses.
y=3[(x1)21]+5y=3[(x-1)^2-1]+5

STEP 6

istribute the3 to both terms inside the parentheses.
y=3(x1)23+5y=3(x-1)^2-3+5

STEP 7

Combine the constants outside the parentheses.
y=3(x1)2+2y=3(x-1)^2+2

STEP 8

Now we have the equation in the form a(xh)2+ka(x-h)^2+k, where a=3a=3, h=1h=1, and k=2k=2.
So, y=3(x1)2+2y=3(x-1)^2+2 is the equation in the form a(xh)2+ka(x-h)^2+k.

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