Lesson 1Find the standard form of a quadraticfunction, and then find the vertex, line ofsymmetry, and maximum or minimumvalue for the defined quadratic function.
The Parabola is defined as "the set of all points P in a plane equidistantfrom a fixed line and a fixed point in the plane." The fixed line is calledthe directrix, and the fixed point is called the focus.A parabola, as shown on the cables of the Golden Gate Bridge(below), can be seen in many different forms. The path that a thrownball takes or the flow of water from a hose each illustrate the shape ofthe parabola.
Each parabola is, in some form, a graph of a second-degree function andhas many properties that are worthy of examination. Lets begin bylooking at the standard form for the equation of a parabola.The standard form is (x - h)2 = 4p (y - k), where the focus is (h, k + p) andthe directrix is y = k - p. If the parabola is rotated so that its vertex is(h,k) and its axis of symmetry is parallel to the x-axis, it has an equationof (y - k)2 = 4p (x - h), where the focus is (h + p, k) and the directrix is x= h - p.It would also be in our best interest to cover another form that the equationof a parabola may appear asy = (x - h)2 + k, where h represents the distance that the parabola hasbeen translated along the x axis, and k represents the distance theparabola has been shifted up and down the y-axis.
Completing the square to get the standard form of a parabola.We should now determine how we will arrive at an equation in the form y = (x - h)2+ k;Example 1Suppose we are given an equation likey = 3x2 + 12x + 1.We now need to complete the square for this equation. I will assume that you havehad some instruction on completing the square; but in case you havent, I will gothrough one example and leave the rest to the reader.When completing the square, we first have to isolate the Ax2 term and the By termfrom the C term. So the first couple of steps will only deal with the first two parts ofthe trinomial.In order to complete the square, the quadratic in the form y = Ax2 + By + C cannothave an A term that is anything other than 1. In our example, A = 3; so we nowneed to divide the 3 out, but that is only out of the 3x2 + 12x terms.This simplifies to y = 3(x2 + 4x) + 1. From here we need to take 1/2 of our B term,then square the product. So in this case, we have 1/2(4) = 2, then 22 is 4. Now,take that 4 and place it inside the parenthetical term.
To update what we have: y = 3(x2 + 4x + 4) + 1; but we now need to keepin mind that we have added a term to our equation that must beaccounted for. By adding 4 to the inside of the parenthesis, we havedone more than just add 4 to the equation. We have now added 4 timesthe 3 that is sitting in front of the parenthetical term. So, really we areadding 12 to the equation, and we must now offset that on the sameside of the equation. We will now offset by subtracting 12 from that 1we left off to the right hand side.To update: y = 3(x2 + 4x + 4) + 1 - 12. We have now successfullycompleted the square. Now we need to get this into more friendlyterms. The inside of the parenthesis (the completed square) can besimplified to (x + 2)2. The final version after the smoke clears is y = 3(x+ 2)2 - 11. And , oh the wealth of information we can pull fromsomething like this! We will find the specifics from this type of equationbelow.
Finding the vertex, line ofsymmetry, and maximum andminimum value for the definedquadratic function.Lets first focus on the secondform mentioned, y =(x - h)2 +k. When we have an equationin this form, we can safely saythat the h represents thesame thing that h representedin the first standard form thatwe mentioned, as does the k.When we have an equation likey = (x - 3)2 + 4, we see that thegraph has been shifted 3 unitsto the right and 4 units upward.The picture below shows thisparabola in the first quadrant.
Had the inside of the parenthesis in the example equation read,"(x+3)" asopposed to "(x-3)," then the graph would have been shifted three units to the leftof the origin. The "+4" at the end of the equation tells the graph to shift up fourunits. Likewise, had the equation read "-4," then the graph would still be pointedupward, but the vertex would have been four units below the x-axis.A great deal can be determined by an equation in this form.
The VertexThe most obvious thing that we can tell, without having to look at the graph, isthe origin. The origin can be found by pairing the h value with the k value, togive the coordinate (h, k). The most obvious mistake that can arise from this isby taking the wrong sign of the h. In our example equation, y = (x - 3)2 + 4,we noticed that the h is 3, but it is often mistaken that the x-coordinate of ourvertex is -3; this is not the case because our standard form for the equation isy = (x - h)2 + k, implying that the we need to change the sign of what is insidethe parenthesis.
The Line of SymmetryTo find the line of symmetry of a parabola in this form, we need to rememberthat we are only dealing with parabolas that are pointed up or down innature. With this in mind, the line of symmetry (also known as the axis ofsymmetry) is the line that splits the parabola into two separate branches thatmirror each other. The line of symmetry goes through the vertex, and sincewe are now only dealing with parabolas that go up and down, the line ofsymmetry must be a vertical line that will begin with "x = _ ". The numberthat goes in this blank will be the x-coordinate of the vertex. For example,when we looked at y = (x - 3)2 + 4, the x-coordinate of the vertex is going be3; so the equation for the line of symmetry is x = 3.
In order to visualize the line of symmetry, take the picture of the parabolaabove and draw an imaginary vertical line through the vertex. If you were totake the equation of that vertical line, you would notice that the line is goingthrough the x-axis at x = 3. An easy mistake that students often make is thatthey say that the line of symmetry is y = 3 since the line is vertical. We mustkeep in mind that the equations for vertical and horizontal lines are the reverseof what you expect them to be. We always say that vertical means "up anddown; so the equation of the line (being parallel to the y-axis) begins with y=__," but we forget that the key is which axis the line goes through. So sincethe line goes through the x-axis, the equation for this vertical line must be x =__.
The Maximum or MinimumIn the line of symmetry discussion, we dealt with the x-coordinate of thevertex; and just like clockwork, we need to now examine the y-coordinate.The y-coordinate of the vertex tells us how high or how low the parabolasits.Once again with our trusty example, y = (x-3)2 + 4, we see that the y-coordinate of the vertex (as derived from the number on the far right of theequation) dictates how high or low on the coordinate plane that theparabola sits. This parabola is resting on the line y = 4 (see line ofsymmetry for why the equation is y = __, instead of x = __ ). Once wehave identified what the y-coordinate is, the last question we have iswhether this number represents a maximum or minimum. We call thisnumber a maximum if the parabola is facing downward (the vertexrepresents the highest point on the parabola), and we can call it aminimum if the parabola is facing upward (the vertex represents thelowest point on the parabola).
How do we tell if the parabola ispointed upward or downward byjust looking at the equation?As long as we have the equation in the form derived from the completing thesquare step, we look and see if there is a negative sign in front of theparenthetical term. If the equation comes in the form of y = - (x - h)2 + k, thenegative in front of the parenthesis tells us that the parabola is pointeddownward (as illustrated in the picture below). If there is no negative sign infront, then the parabola faces upward.