Quadratic Equations in Vertex Form

In summary, the roots of 3x^2 + 5x - 2 are x = 1/3 and x = -2. However, when finding the vertex, there was an error in completing the square. The correct vertex is (-5/6, -4 1/12).
  • #1
alex2256
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Homework Statement


Find the roots and the vertex for 3x^2 + 5x - 2


Homework Equations


ax^2 + bx + c
a(x - h) + k

The Attempt at a Solution


OK, this is what I attempted.

3x^2 + 5x - 2 = 0
3x^2 + 6x - x - 2 = 0
3x(x+2) - 1(x+2) = 0
x = 1/3 and x = -2

OK, so I found the roots, however, I am having difficulty understanding finding the vertex of THIS problem. The book says the vertex should be (-5/6, -4 1/12), so here's what I did.

3x^2 + 5x - 2 = 0
x^2 + 5/3x - 2/3 = 0
x^2 + 5/3x - (25/36) - (2/3 - 25/35) = 0
(x + 5/6)^2 - 49/36

therefore the vertex is -5/6 and -49/36; obviously I didn't get -4 1/12.

Can somebody help me and explain to me what's going on?
 
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  • #2
Yup, it's pretty simple. You went from

x2+5/3x-25/36 (emphasis mine EDIT: Actually, a bold minus sign looks no different than a regular one...)

to

(x+5/6)2

if yo factor this out, you'll see you get a +25/36. So your constant term is off because you put the negative 25/36 into the complete the square portion instead of the positive part
 
  • #3


Firstly, it is important to note that the vertex form of a quadratic equation is given by a(x-h)^2 + k, where (h,k) is the vertex. In your attempt at a solution, you have correctly found the roots of the equation, but your method for finding the vertex is incorrect.

To find the vertex, we can use the formula h = -b/2a, where a and b are the coefficients of x^2 and x respectively. In this case, a = 3 and b = 5, so h = -5/6. Now, to find k, we can substitute the value of h into the equation and solve for k.

3(-5/6)^2 + 5(-5/6) - 2 = k
25/12 - 25/6 - 2 = k
-49/12 = k

Therefore, the vertex is (-5/6, -49/12), which can also be written as -4 1/12.

In your attempt, you have correctly completed the square, but you have made a mistake in the step where you subtracted 2/3 - 25/36. The correct expression should be (2/3)^2 - 25/36, which simplifies to -49/36. This is why you ended up with -49/36 instead of -4 1/12 as the vertex.

I hope this helps clarify the process of finding the vertex of a quadratic equation in vertex form. Remember to always check your calculations and to use the correct formula for finding the vertex.
 

FAQ: Quadratic Equations in Vertex Form

1. What is a quadratic equation in vertex form?

A quadratic equation in vertex form is a way of expressing a quadratic equation in the form y = a(x-h)^2 + k, where (h,k) represents the coordinates of the vertex of the parabola. This form is useful for graphing and finding important features of the parabola such as the vertex, axis of symmetry, and direction of opening.

2. How do I convert a quadratic equation from standard form to vertex form?

To convert a quadratic equation from standard form (y = ax^2 + bx + c) to vertex form, you can use the process of completing the square. This involves finding the value of h and k in y = a(x-h)^2 + k. The value of h can be found by taking half of the coefficient of x (b) and squaring it, while the value of k can be found by adding and subtracting the square of b/2 from the constant term (c).

3. What is the significance of the vertex in a quadratic equation?

The vertex in a quadratic equation is the point where the parabola changes direction and is the highest or lowest point on the graph. It is also the point where the axis of symmetry intersects the parabola. The coordinates of the vertex (h,k) can provide important information about the parabola, such as its minimum or maximum value and the direction in which it opens.

4. Can a quadratic equation have more than one vertex?

No, a quadratic equation can only have one vertex. This is because the vertex represents the highest or lowest point on the parabola, and there can only be one highest or lowest point. However, a quadratic equation can have more than one x-intercept (where the parabola crosses the x-axis) and may have no x-intercepts if the parabola does not intersect the x-axis.

5. How can I use the vertex form of a quadratic equation to solve real-world problems?

The vertex form of a quadratic equation is useful for solving real-world problems that involve finding maximum or minimum values. For example, it can be used to model the trajectory of a projectile, where the vertex represents the highest point the projectile reaches. It can also be used to determine the optimal production level for a business, where the vertex represents the maximum profit. By finding the coordinates of the vertex, you can determine the optimal solution to the problem at hand.

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