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The equation y=x^(2)+6x+b is graphed in the xy-plane. If the vertex of the graph of the equation is at (-3,0), what is the value of b ?

The equation y=x2+6x+b y=x^{2}+6 x+b is graphed in the xy x y -plane. If the vertex of the graph of the equation is at (−3,0) (-3,0) , what is the value of b b ?

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Q. The equation y=x2+6x+b y=x^{2}+6 x+b is graphed in the xy x y -plane. If the vertex of the graph of the equation is at (−3,0) (-3,0) , what is the value of b b ?
  1. Identify vertex form: Identify the vertex form of a parabola.\newlineThe vertex form of a parabola is y=a(x−h)2+ky = a(x - h)^2 + k, where (h,k)(h, k) is the vertex of the parabola.
  2. Write equation with given vertex: Use the given vertex (−3,0)(-3, 0) to write the equation in vertex form.\newlineSince the vertex is (−3,0)(-3, 0), we have h=−3h = -3 and k=0k = 0. Thus, the vertex form of the equation is y=a(x−(−3))2+0y = a(x - (-3))^2 + 0, which simplifies to y=a(x+3)2y = a(x + 3)^2.
  3. Expand vertex form: Expand the vertex form to compare it with the given equation.\newlineExpanding y=a(x+3)2y = a(x + 3)^2 gives us y=a(x2+6x+9)y = a(x^2 + 6x + 9). Since the given equation is y=x2+6x+by = x^2 + 6x + b, we can compare the coefficients to find the value of aa and bb.
  4. Compare coefficients for aa: Compare the coefficients of x2x^2 and xx to find the value of aa.\newlineThe coefficient of x2x^2 in both the expanded vertex form and the given equation is 11, so a=1a = 1. The coefficient of xx is also the same (66), which confirms that a=1a = 1 is correct.
  5. Determine value of bb: Determine the value of bb by comparing the constant terms.\newlineIn the expanded vertex form, the constant term is 9a9a, which is 99 since a=1a = 1. For the vertex form to match the given equation, the constant term must also be bb. Therefore, b=9b = 9.
  6. Verify vertex with b=9 b = 9 : Verify that the vertex (−3,0) (-3, 0) is correct with b=9 b = 9 .\newlineSubstitute x=−3 x = -3 into the given equation y=x2+6x+9 y = x^2 + 6x + 9 to check if y y equals 0 0 . y=(−3)2+6(−3)+9=9−18+9=0 y = (-3)^2 + 6(-3) + 9 = 9 - 18 + 9 = 0 . This confirms that the vertex is indeed (−3,0) (-3, 0) when b=9 b = 9 .

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