Write the equation in vertex form.
y = x² + 6x + 9
Vertex form of a quadratic equation is y = a(x - h) + k,
where h and k are the x and y coordinate of the vertex. One Method Find the x-coordinate of the vertex, which is x = -b/2a In y = x² + 6x + 9, a = 1 {the coefficient of x²} b = 6 {the coefficient of x} c = 9 {the constant term} x = -b/2a {the x-coordinate of the vertex} = -6/2(1) {substituted 6 in for b and 1 in for a} x = -3 {divided} the x-coordinate of the vertex is -3 Then to find the y-coordinate, substitute -3 back in for x. y = (-3)² + 6(-3) + 9 {substituted -3, in for x, into y = x² + 6x + 9} y = 9 - 18 + 9 {evaluated exponent and multiplied} y = 0 {subtracted and added} the y-coordinate of the vertex is 0 Substitute the x and y coordinates back into vertex form, along with a, a is the same in vertex form as it is in standard form {the original equation was in standard form}. y = a(x - h)² + k {vertex form of a quadratic equation} y = [x - (-3)]² + 0 {substituted 1 in for a, -3 in for x, and 0 in for k, into vertex form} y = (x + 3)² {simplified} Another Method Complete the square within the equation: y = x² + 6x + 9 Take 1/2 of the coefficient of x, square it and add it to the first two terms, also subtract it from the 9 to keep the equation balanced. y = x² + 6x + 9 + 9 - 9 {added 9 and subtracted 9 from the right side} y = x² + 6x + 9 {simplified} y = (x + 3)² {factored the binomial, which you could've done from the very beginning!} Ask Algebra House
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