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How to convert y=2(x-3) (x+5) to vertex form?

User Pmakholm
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2 Answers

1 vote

Step-by-step explanation:

y = 2(x-3) (x+5)

= 2(x²+2x-15)

= 2x²+4x-30

= 2x²+4x+2-32

= 2(x²+2x+1)-32

y = 2(x+1)² -32

hope this helps

User Agent Biscutt
by
8.7k points
1 vote

Final answer:

To convert y=2(x-3)(x+5) to vertex form, expand and simplify the equation, factor out the x^2 coefficient, complete the square by adding and subtracting (b/2)^2, rewrite the perfect square trinomial, distribute the coefficient and combine constants. The resulting vertex form is y=2(x+1)^2-32, with vertex (-1, -32).

Step-by-step explanation:

To convert the quadratic equation y = 2(x - 3)(x + 5) to vertex form, we must complete the square. The vertex form of a quadratic equation is y = a(x - h)^2 + k, where (h, k) is the vertex of the parabola.

  1. First, expand the given equation:
    y = 2(x^2 + 5x - 3x - 15)
  2. Simplify the terms:
    y = 2(x^2 + 2x - 15)
  3. Factor out the coefficient of the x^2 term within the parentheses:
    y = 2[(x^2 + 2x) - 15]
  4. Add and subtract (b/2)^2 inside the brackets, where b is the coefficient of x:
    y = 2[(x^2 + 2x + (1)^2) - 1^2 - 15]
  5. Rewrite the perfect square trinomial:
    y = 2[(x + 1)^2 - 16]
  6. Distribute the 2 and combine the constants:
    y = 2(x + 1)^2 - 32

Now, the equation is in vertex form, which is y = 2(x + 1)^2 - 32. The vertex of this parabola is (-1, -32).

User Evert Heylen
by
8.1k points

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