Write quadratic function in standard (vertex) form.?

Question

f(x)= -X^2+3X-1 

and 

f(x)= 5X^2+10X-1 

Please help! Thank you

Answer 1

The function f(x) = - x² + 3x - 1

y = - x² + 3x - 1

Vertex form of a parabola is y = a(x - h)² + k.

Here x² coefficient is - 1, for perfect square make x² coefficient 1 by multiply each side by negative one.

- y =  x² - 3x + 1

To change the expression into a perfect square trinomial add (half the x coefficient)² to each side of the expression.

x  coefficient = - 3 then (half the x coefficient)² is (- 3/2)² = 9/4.

So, Add 9/4 to each side.

- y + (9/4) = x² - 3x + 1 + (9/4)

- y + (9/4) = [x - (3/2)]² + 1

 y =(- 1){[x - (3/2)]² + 1 - (9/4)}

y = (- 1)[x - (3/2)]² + (5/4)

Therefore, vertex form of the parabola is y = (- 1)[x - (3/2)]² + (5/4).

Answer 2

The function f(x) = 5x² + 10x - 1

Vertex form of a parabola is y = a(x - h)² + k.

y = 5x² + 10x - 1

Take out common factor.

y = 5[x² + 2x - (1/5)]

To change the expression (x² + 2x) into a perfect square trinomial add (half the x coefficient)² to each side of the expression.

 Here x coefficient = 2. So, (half the x coefficient)² = (2/2)² = 1.

Add and subtract 1 to the expression.

y = 5[x² + 2x - (1/5) + 1 - 1]

y = 5[x² + 2x + 1 - (1/5) - 1]

y = 5[x² + 2x + 1 - (6/5)]

y =5(x² + 2(1)(x) + 1²) - 6

Apply Perfect Square Trinomial : u² + 2uv + v² = (u + v)².

y = 5(x + 1)² - 6

Vertex form is y = 5[x - (- 1)]² +(- 6).