How do I convert the equation $f (x) = x^{2} - 4 x + 3$ $f(x)=x^2-4x+3$ to vertex form?

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Answer 1 · 2015-07-29T19:16:53

$f (x) = {(x - 2)}^{2} - 1$ $color(red)( f(x) = (x-2)^2-1)$

The vertex form of a quadratic is given by $y = a(x – h)^2 + k$ , where ( $h, k$ ) is the vertex.

The " $a$ " in the vertex form is the same " $a$ " as in $y = ax^2 + bx + c$ .

Your equation is

$f(x) = x^2-4x+3$

We convert to the "vertex form" by completing the square.

Step 1. Move the constant to the other side.

$f(x)-3 = x^2-4x$

Step 2. Square the coefficient of $x$ and divide by 4.

$(-4)^2/4 = 16/4 = 4$

Step 3. Add this value to each side

$f(x)-3+4= x^2-4x+4$

Step 4. Combine terms.

$f(x)+1 = x^2-4x+4$

Step 5. Express the right hand side as a square.

$f(x)+1= (x-2)^2$

Step 5. Isolate $f(x)$ .

$f(x) = (x-2)^2-1$

The equation is now in vertex form.

$y = a(x – h)^2 + k$ , where ( $h, k$ ) is the vertex.

$h = 2$ and $k = -1$ , so the vertex is at ( $2,-1$ ).

Graph

How do I convert the equation f(x)=x^2-4x+3f(x)=x2−4x+3f(x)=x^2-4x+3 to vertex form?