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Converting standard form into vertex form and vice-versa?

I have a test tomorrow, and I was wondering how do you convert an equation in standard form into vertex form and vice-versa ?

I’m guessing that you mean this question in regard to parabolas. Thus, given
y = ax² + bx + c
how do we convert it to the form
y – k = a(x – h)² ?
Move the constant term from the right to the left side, then complete the square on the right.
e.g. y = x² – 10x + 28
y – 28 = x² – 10x
Divide -10 by 2, and square the result. Gives 25, so add 25 to both sides:
y – 3 = x² – 10x + 25
y – 3 = (x – 5)² so the vertex is (5, 3)

Again e.g. y = 3x² + 7x + 5
y – 5 = 3x² + 7x
y – 5 = 3(x² + (7/3)x )
Divide 7/3 by 2, result 7/6, square it gives 49/36 so add 3*49/36 to both sides.
y – 5 + 49/12 = 3(x² + (7/3)x + 49/36)
y – 11/12 = 3(x + 7/6)²
hence the vertex is (-7/6, 11/12)

SORRY, I overlooked the “vice-versa”.
Given y – k = a(x – h)²
just expand the right-hand side and make y the subject of the equation.
If given the vertex and one other point on the parabola, substitute the coordinates of the vertex; then in the equation the only unknown constant is a. To find it, substitute the coordinates of the other point on the parabola for x and y.
e.g. Find in standard form the equation of the parabola with vertex (4, -1) passing through the point (2, -3)
First, using h = 4, k = -1:
y + 1 = a(x – 4)²
Now substitute x = 2, y = -3:
-3 + 1 = a(2-4)²
-2 = 4a
a = -(1/2)
Hence the equation is
y + 1 = -(1/2)(x – 4)²
y + 1 = -(1/2)x² + 4x – 8
y = -(1/2)x² + 4x – 9

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How To Solve Quadratic Equations – Home Tutoring Services

Before tackling the issue of how to solve a quadratic equation, it is important to be able to identify one!

A quadratic function is any function in which the power of the leading coefficient is equal to two.

That is, the highest power of the independent variable in the function has to be two. The standard/general form of such a function is f(x) = ax^2 + bx + c, where a, b, c are constants and a cannot be equal to zero. When a > 0, the graph of this function is a parabola that opens upward and when a < 0, the graph represents a downward-opening parabola.

A quadratic function may also be expressed in vertex form as f(x) = a(x – h)^2 + k, where a cannot equal zero and the vertex of the graph is at (h, k). As is the case with the standard form of the quadratic function, if a > 0, the graph opens upward and if a < 0, the graph opens downward.

Many times, the quadratic function is set equal to zero (with f(x) = 0) in order to solve for the x-intercepts (or roots) of the function. This process may also be referred to as solving a quadratic equation. When this is done, the quadratic function in standard form becomes a quadratic equation of the form ax^2 + bx + c = 0.

The first method that can be used to solve a quadratic equation is factoring:

The quadratic equation in standard form can be factored into two binomials, that is, two polynomials with two terms. For example, 2x^2 + x – 3 = 0 can be factored into (2x + 3)(x – 1) = 0.

Once this is done, each of these binomials can be solved for x. These values of x represent the x-intercepts.

One drawback to this method is that not every quadratic equation can be factored. Also, some factorable quadratic equations are not very easy to factor.

2.         The second method that can be used is the quadratic formula:

The equation must be converted to standard form in order to use the formula. This formula can be found in any Precalculus textbook or even on the internet.

The advantage of using the formula is that it can be used even when the quadratic equation is not factorable. Also, in the case that there are no real solutions for the equation, imaginary solutions can easily be determined.

A minor disadvantage of using this equation is that it is possible to make an error in calculating the solutions if a wrong number is plugged into the formula or a negative is neglected in the process of making calculations.

A third method that can be used to solve a quadratic equation is completing the square:

Though this method is not difficult per say, there are many instances in which computational errors can be made. The good thing about completing the square is that is can also be used to convert a quadratic function from standard or general form to vertex form.

Keep in mind that there are many advantages to having a quadratic equation in vertex form.

The fourth method used to solve a quadratic equation is the graphing calculator method:

The original quadratic function can be entered into the calculator either in standard or vertex form. Once this is done, the calculator’s calculation capabilities can be used to find the roots of the function.

On most of the TI calculators, the quadratic function can be entered once the “Y =” button is pressed. After that, the graphing window may have to be adjusted in order to see the complete graph.

The next step would involve pressing the “2nd” and “TRACE” buttons simultaneously in order to access the CALC menu. Under the CALC menu, the “root” option should be selected in order to determine the roots.

Note that each root has to be determined separately.

In conclusion, there are four options that can be used to solve a quadratic equation: factoring, the quadratic formula, completing the square, and the graphing calculator method. They should each be known so that they can be used interchangeably.

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