What are the vertexes for each problem?

Question

Y=x^2+6x+8 Y=x^2-2x+3 These problems are both in standard form.?

Answer 1

The parabola equation is y = x² + 6x + 8

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

To change the expression (x² + 6x) into a perfect square trinomial add (half the x coefficient)² to each side of the expression.

 Here x coefficient = 6. So, (half the x coefficient)² = (6/2)² = 9.

Add and subtract 9 to the expression.

y = (x² + 6x + 8 + 9 - 9)

y = (x² + 6x + 9 + 8 - 9)

y = (x² + 6x + 9 - 1)

y = (x² + 2(3)(x) + 3²) - 1

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

y = (x + 3)² - 1

The parabola in vertex form y = [x - (-3)]² + (- 1)

Vertex (h, k) = (- 3, - 1)

Answer 2

The parabola equation is y = x² - 2x + 3

Compare it to standard form of parabola y = ax² + bx + c

a = 1, b = - 2, c = 3

To find the x coordinate of vertex, first find the axis of symmetry x = - b/2a

x = - (- 2 )/2(1)

x = 1

To find y coordinate of vertex substitute the x = 1 in y = x² - 2x + 3.

y = (1)² - 2(1) + 3

y = 1 - 2 + 3

y = 2

Vertex is (x , y) = (1, 2).