#Quadratic Functions Examples How To Calculate Them

In the first example, we will graph the quadratic function. Simply, the three terms include one that hasan x 2, one has an x, and one term is 'by itself'with no x 2 or x.We call this graphing quadratic functions using transformations. Normally, we see thestandard quadratic equation written as the sum of three termsset equal to zero. So, for our purposes, we willbe working with quadratic equations which mean that the highestdegree we'll be encountering is a square. In an algebraic sense, the definition ofsomething quadratic involves the square and no higher power ofan unknown quantity second degree. Similarly, one of the definitions of the termquadratic is a square.

Note thatif a = 0, the x 2 term would disappear and we wouldhave a linear equation!The solutions to quadratic equations are called roots. Thus, the standardized form of a quadratic equation is ax 2+ bx + c = 0, where 'a' does not equal 0. When a quadratic function is in standard form, then it is easy to sketch its graph by reflecting, shifting, and stretching/shrinking the parabola y x 2. F ( x) x 2 + k.The functions in parts (a) and (b) of Exercise 1 are examples of quadratic functions in standard form. Then we will see what effect adding a constant, k, to the equation will have on the graph of the new function.

X 2 + 5 x + 4 0, the related quadratic function is. For example, if you are given the quadratic equation. For every quadratic equation, there is a related quadratic function.

If youdraw a vertical line through the vertex, it will split the parabolain half so that either side of the vertical line is symmetricwith respect to the other side.This vertical line is called the line of symmetry or axis of symmetry. The lowest pointon the graph is (1, -1) and is called the vertex. Let's substitute the same values in for x as we did in thechart above and see what we get for y.What are the x-and y-intercepts? What is the lowest point onthe graph?Here, we see again that the x- and y-intercepts are both (0,0), as the parabola crosses through the origin. So, we would have the equation, y = x 2-2x. The secondform is called the vertex-form or the a-h-k form,y = a(x - h) 2 + k.Parabolas in the standard from y = ax 2 + bx +c.Let's trying graphing another parabola where a = 1, b = -2and c = 0.

So our y-intercept is (0, 1). The actual point of they-intercept is (0, y), so x is always 0.If substitute 0 in for x, we'll get y = 1 as indicated in thechart above. Substitute our standard values in for x and solve for yas illustrated in the chart below:The points and the graph through these points are shown below.Graph of the quadratic function y = -3x 2+ x + 1What is the y-intercept? Can you estimate the x-intercepts?Can you estimate the vertex? What is the general shape of theparabola?Remember, to find the y-intercept of any equation, we can alwayssubstitute 0 in for x and solve for y. So, in this case, the lineof symmetry would be x = 1.The vertex is the lowest point on the parabola if the parabolaopens upward and is the highest point on the parabola ifthe parabola opens downward.Now let's try graphing the parabola: y = -3x 2 +x + 1. Remember, to graph a vertical line, go across thex-axis to the value of 'c' where the equation indicates,x = c, and draw the vertical line. To find the equation of the line of symmetry,it will always be y = c, where c is always the x-value of thevertex (x, y).

Quadratic Functions Examples How To Calculate Them

We would solve for the values ofx using the quadratic formula. If we substituted 0 in for y, we would get the equation0 = -3x 2 + x + 1. We'll estimate them now, aswe will find out how to calculate them in detail in the next lesson,'The Quadratic Formula.'For now, remember that you would solve for the x-interceptsby substituting 0 in for y and solving for x, as you would forany equation.

But what is the vertex exactly?The vertex is an important coordinate to find because we knowthat the graph of the parabola is symmetric with respect to thevertical line passing through the vertex. It looks like thex-value of the vertex is a little less than 1/4 of the way fromthe origin to x = 1, and the y-value of the vertex is a littlemore than 1. If you don't know thequadratic formula, not to worry, you're not supposed to! We'llcome back to this equation in detail in the next lesson.What about the vertex? You can't really tell the exact valueof the vertex just by looking at the graph.

So y = -11The vertex for this parabola would be (3, -11).Practice graphing the equation by plotting the vertex and they-intercept as shown below. So our x-value of the vertex is x = 3.To find the y-value, we would substitute 3 in for x, or findf(3) = 2(3) 2 12(3) + 7 = -11. We do this justas we have done all along by substituting 1/6 in for x in ourequation and solving for y.So, we'll substitute 1/6 in for x in our original equation,y = -3x 2 + x + 1Where 13/12 is equal to 1 and 1/12, which is slightly greaterthan 1.So our vertex is (1/6, 13/12), which would probably be hardto find just by looking at the graph!Notice that the function increases from negative infinity tox = 1/6 (the x-value of the vertex), and then decreases from x= 1/6 to positive infinity along the x-axis.Find the vertex of the parabola y = 2x 2 - 12x +7.To find the x-value of the vertex x = -b/2a , we would substitute-12 in for 'b' and 2 in for 'a', to get x= - (-12)/(2(2)) = 3. Inthis equation, 'b' is the coefficient of the x-termand 'a', like always, is the coefficient of the x 2term.So now we have the x-value of the vertex, x = 1/6, so our vertexso far is in the form (1/6, y).Now, all we need to do is to find the y-value.

One of the general characteristics of aparabola is that if the value of 'a' which isthe coefficient of the x 2 term is negative, theparabola will open downward. Also, anotherpoint is (2, -9).Plot this function and make sure your graph looks similar tothe one shown below.The Value of 'a,' the Coefficient of the x 2Term.In most of the previous examples, the parabola opened upward.However, the parabola with the equation y = -3x 2 +x + 1 opened downward. So, another point on this graph is (1, 6). So, plot the vertex, the y-intercept (0, 2).What would the y-value of a coordinate be if the x-value =1? Substitute 1 in for x in this equation, and you would get y= 6. Remember, you can pick any number to substitute in the equationfor x and solve for y, and the corresponding point will be onthe graph.

On the other hand, if 'a'is negative, the graph opens downward and the vertex is the maximumvalue.Now, let's refer back to our original graph, y = x 2,where 'a' is 1.In this equation, when we substitute values in for x and solvefor y, we just square x as we did above. Knowing the characteristics of the graph beforeyou start graphing is a great checking devise!So, given a quadratic function, y = ax 2 + bx + c,when 'a' is positive, the parabola opens upward andthe vertex is the minimum value. So, if you aregraphing a parabola and it has '- a' for the x 2coefficient, and the graph ends up opening upward, you know thatyou must be graphing incorrectly! This is no problem, just checkyour calculations.