Graphing Quadratic Functions - Pre-Calculus

To solve for the vertex form, we must start by completing the square:

This process outlines how to convert a quadratic function to vertex form:

To get the equation into vertex form, we factor the largest constant from the terms with a degree of greater than or equal to 1.

We then complete the square by following these steps

1. finding half of the coefficients of the term
2. squaring that result
3. and then adding that square to the expression for .

Keep in mind that what is done on one side of the equation must be done on the other.

And factoring the quadratic polynomial of x

we get

All quadratic functions have a vertex and many cross the x axis at points called zeros or roots. If we know the vertex and its zeros, quadratic functions become very easy to draw since the vertex is also a line of symmetry (the zeros are equidistant from the vertex on either side).

Factor the equation to get and . Thus, the roots are 3 and -2.

The vertex can be found by using .

simplify

.

The axis of symmetry is halfway between the two roots, or simply the x coordinate of the vertex. So the axis of symmetry lies on x=1/2. To graph, draw a point at the coordinate pair of the vertex. Then draw points on the x axis at the roots, and finally, trace upwards from the vertex through the roots with a gentle curve.

Finding the vertex, intercept and axis of symmetry are crucial to finding the function that corresponds to the graph:

The vertex form of a quadratic function is written as:

and the coordinates for the vertex are:

Looking at the graph and the position of the axis of symmetry, the vertex is positioned at , leaving us with an equation so far of:

While we don't know a right away, is the only option that really works. The y-intercept is at and we can plug that into the formula to confirm that this is the correct function:

Immediately we can tell that the equation has a negative coefficient, because the parabola opens downward, forming an umbrella shape. Based upon the information given in the figure, we can use the intercepts, axis of symmetry, and the vertex to identify the equation of the parabola. Let's observe the vertex form of a parabola written as the following:

In this equation, is the vertex of the parabola, and determines whether the parabola opens upwards or downwards. The axis of symmetry is at and the vertex is located at , which we can plug into the following function:

We know that is negative because of the position of the parabola.

To find the vertex of a quadratic equation, you'll look to put the quadratic in the form , where is then the vertex. To get from the original equation to vertex form, you'll have to complete the square by looking at the terms that include and to turn that into a perfect square. Here you should see that with as the first two terms, you could have a perfect square if you could use . So to complete the square, you can express the given quadratic as:

Note that the +1 and -10 terms net out to the -9 that was in the original equation, so in this case you have not altered the value at all, but have merely reallocated numbers to fit vertex form (also note that there was no coefficient to the term, making the term in vertex form equal to 1).

From here you can factor the quadratic on the left to perfectly math vertex form:

This means that and , making the vertex .

Note that solving for the x-coordinate of the vertex of a parabola also tells you its line of symmetry, so your job here is to put the quadratic into Vertex Form in order to find the vertex, which will give you the line of symmetry. Vertex form is , where is then the vertex. To get to that form, you will want to complete the square by looking at the and terms and determining which perfect square equation they belong to. To do that, separate those two terms from the -2 term, and then factor out the coefficient of 3:

Then note that the way to turn into a perfect square would be to add 1 to it to get to . Of course, you cannot just add one within the parentheses without balancing the rest of the equation on both sides. Since that +1 will be multiplied by a coefficient of 3, you should add 3 to the right side of the equation to match what you've done on the left:

Then you can factor the quadratic on the left into perfect square form, and subtract 3 from both sides to reset to 0:

This provides you with Vertex Form, so you can say that and , making the vertex and the Line of Symmetry just the x-coordinate of .

The line of symmetry of a parabola is the x-coordinate of its vertex, so you can solve this problem by taking the given quadratic and converting to Vertex Form, , where is the vertex. To do so, focus on the and terms first, pulling them aside and factoring out the common 4 so that you have your coefficient:

Next, think of which Perfect Square quadratic you can form using the terms in parentheses. , so if you add 1 within the parentheses you can treat it as a perfect square to match Vertex Form. Of course, you can't just add 1 -- which will be multiplied by the coefficient of 4 -- without accounting for it on the other side of the equation. So as you transform the parentheses on the left to match Vertex Form and add the +1 within parentheses to do so, also add 4 to the right hand side to stay balanced:

Now you can factor the quadratic to Perfect Square form, and subtract 4 from both sides to finish Vertex Form:

This means that and , so the line of symmetry - which is the x-coordinate - is at .