Quadratic function is every function that has the form \(f(x) = ax^2 + bx + c,\) for \(a,b,c\in \mathbb{R}\) and \(a \neq 0.\) The domain of quadratic function are all real numbers. A graph of a quadratic function is a curve called parabola. In Figure 1 the function \(f(x) = x^2\) is shown.
Figure 1 - The graph of a quadratic function \(f(x) = x^2\)
The quadratic function \(f(x) = ax^2 \) is even function (\(f(-x) = f(x)\)), it is symmetrical with respect to y-axis. The parabola can be classified by concavity and by the number of roots. If coefficient \(a>0\) than the parabola is Concave up which means that parabola opens upward. If \(a < 0\) than the parabola is concave down which means that parabola opens downwards. The concave up and concave down types of quadratic function i.e parabola is shown in the following figure. FIGURE MISSING
The other classification is by the number of roots i.e. the number of times the parabola itersects with the axis line. If the parabola has no roots it means that parabola do not touches the \(x\) and \(y\) axis. If the parabola has one root than the parabola vertex touches \(x\) axis and if the parabola has two roots the parabola goes through the \(x\) axis twice. For better understanding each case is shown in following figure.
The point at which the parabola with positive leading coefficient is lowest and the with negative leading coefficient is the highest is called the vertex of parabola. For example the vertex of parabola \(f(x) = ax^2 \) is at the origin of coordinate system with point \(T(0,0).\) The size of the parabola \(f(x) = ax^2\) changes as the value of leading coefficient changes. If the value \(|a|\) is lower the parabola is wider and if the vlaue of \(|a|\) is higher the parabola is narrower.
The parabola \(f(x) = x^2 \) and \(f(x) = ax^2 + bx + c\) have the same shape. The graph of parabola \(f(x) = ax^2 + bx +c\) is achieved with translation of parabola \(f(x) = ax^2\) for \(x_0\) in direction of \(x\) axis and \(y_0\) in direction of the \(y\) axis where the \(x_0\) and \(y_0\) are vertex coordinates of translated parabola and are calculated using formulas which can be written as: \begin{eqnarray} x_0 &=& -\frac{b}{2a},\\ y_0 &=& \frac{4ac - b^2 }{4a}\\ T&&(x_0,y_0) \end{eqnarray} The parabola \(f(x) = ax^2 + bx +c\) can be written in following form: \begin{eqnarray} y &=& a(x-x_0)^2 + y_0 \end{eqnarray} The value of quadratic function \(y_0 = f(x_0)\) is called extreme value of the function where \(x_0\) and \(y_0\) are vertex coordinates. The extreme value of the function can be minimum and maximum. The function achieves the minimum value if \(a\) is greater than zero \(a>0\). The function achieves maximum value if \(a\) is lower than zero \(a < 0 \).
Every parabola is symmetric with respect to axis of symmetry. The axis of symmetry of parabola \(f(x) = ax^2 + bx +c \) is a line which is perpendicular on \(x\) axis and goes through the vertex of parabola.
Figure 1 - The graph of a quadratic function \(f(x) = x^2\)
The quadratic function \(f(x) = ax^2 \) is even function (\(f(-x) = f(x)\)), it is symmetrical with respect to y-axis. The parabola can be classified by concavity and by the number of roots. If coefficient \(a>0\) than the parabola is Concave up which means that parabola opens upward. If \(a < 0\) than the parabola is concave down which means that parabola opens downwards. The concave up and concave down types of quadratic function i.e parabola is shown in the following figure. FIGURE MISSING
The point at which the parabola with positive leading coefficient is lowest and the with negative leading coefficient is the highest is called the vertex of parabola. For example the vertex of parabola \(f(x) = ax^2 \) is at the origin of coordinate system with point \(T(0,0).\) The size of the parabola \(f(x) = ax^2\) changes as the value of leading coefficient changes. If the value \(|a|\) is lower the parabola is wider and if the vlaue of \(|a|\) is higher the parabola is narrower.
The parabola \(f(x) = x^2 \) and \(f(x) = ax^2 + bx + c\) have the same shape. The graph of parabola \(f(x) = ax^2 + bx +c\) is achieved with translation of parabola \(f(x) = ax^2\) for \(x_0\) in direction of \(x\) axis and \(y_0\) in direction of the \(y\) axis where the \(x_0\) and \(y_0\) are vertex coordinates of translated parabola and are calculated using formulas which can be written as: \begin{eqnarray} x_0 &=& -\frac{b}{2a},\\ y_0 &=& \frac{4ac - b^2 }{4a}\\ T&&(x_0,y_0) \end{eqnarray} The parabola \(f(x) = ax^2 + bx +c\) can be written in following form: \begin{eqnarray} y &=& a(x-x_0)^2 + y_0 \end{eqnarray} The value of quadratic function \(y_0 = f(x_0)\) is called extreme value of the function where \(x_0\) and \(y_0\) are vertex coordinates. The extreme value of the function can be minimum and maximum. The function achieves the minimum value if \(a\) is greater than zero \(a>0\). The function achieves maximum value if \(a\) is lower than zero \(a < 0 \).
Every parabola is symmetric with respect to axis of symmetry. The axis of symmetry of parabola \(f(x) = ax^2 + bx +c \) is a line which is perpendicular on \(x\) axis and goes through the vertex of parabola.
- Example 1 - Determine the equation of parabola that has vertex at point \(\mathrm{T}\left(-1,3\right)\) and passes through point \(\mathrm{A}\left(2,-1)\right)\)
Solution - $$\begin{eqnarray}y &=& a(x-x_0)^2+y_0\\\nonumber y &=& a(x-(-1))^2 +3\\\nonumber y&=& a(x+1)^2 + 3 \\ \nonumber y&=& a(x^2+2x+1)+3 \\ \nonumber y&=& ax^2+2ax+a +3\end{eqnarray}$$ Using point A the value of a can be determined. $$\begin{eqnarray}f(2)&=& a\cdot2^2+2\cdot a \cdot 2 + a + 3 \\\nonumber 4a +4a +a +3 &=&-1 \\\nonumber 9a &=& -4 \\\nonumber a &=& -\frac{4}{9}\end{eqnarray}$$ The equation of parabola can be written in simplified from $$y = - \frac{4}{9}(x+1)^2 +3 $$ or in expanded form $$ y = -\frac{4}{9}x^2 -\frac{8}{9}x + \frac{23}{9}$$ - Example 2 Determine the parabola equations that is shown in following figure and goes through the points \(A(-1,3)\) and \(B(2,-1).\)
Solution - The general equation of the parabola can be written as \(y = a(x-x_0)^2 + y_0\) where \(x_0\) and \(y_0\) represent the coordinates of parabola vertex respectively. Our parabola has the vertex which is also a maximum of the parabola at point \(A(-1,3)\) So, from point A the \(x_0\) and \(y_0\) can be obtained i.e \(x_0 = -1\) and \(y_0 = 3\). By inserting these values into general parabola equation the following solution is obtained. \begin{eqnarray}y &=& a(x-x_0)^2 + y_0 \\ y &=& a(x-(-1))^2 + 3 = a(x+1)^2 + 3\end{eqnarray} The only thing that must be determined is coefficient \(a\) which will be determined by inserting the coordinates of point B into the parabola equation. \begin{eqnarray} y = a(x+1)^2 +3 \\ a(2+1)^2 + 3 &=& -1 \\9a &=& -4/:9\Rightarrow a &=& -\frac{4}{9}\end{eqnarray} The quadratic function can be written as \(y = -\frac{4}{9}x^2 - \frac{8}{9}x + \frac{23}{9}\) - Example 3 Determine the equation of quadratic function \(f(x) = ax^2 + bx + c\) if \(f(-1) = f(3) = 0\), and \(f(1) = 2\).
Solution From the given information it can easily be concluded that \(f(-1) = f(3) = 0\) represent the roots of the qadratic function. The axis of symmetry of parabola has an equal distance from both roots and passes through the point that is half of the lenght between two roots. From that statement the vertex coordinate can be calculated as: $$ x_0 = \frac{-1 + 3}{2} = \frac{2}{2} = 1.$$ The other given information about quadratic function is that for \(x = 1\) the value of the quadratic founction is 2 (\(f(1) = 2\)). So, we can see that the third value of the function actually represent the vertex of the parabola (quadratic function) \(T(1,2).\) Using fomrula \(f(x) = a(x-x_0)^2 + 2\) and the calculated values so far the quadratic function can be determine. \begin{eqnarray} T(1,2) &\Rightarrow& x_0 = 1, y_0 = 2\\ f(x) &=& a(x-1)^2 +2 \Rightarrow a = ? \\ f(3) &=& 0 \Rightarrow a(3-1)^2 + 2 = 0\\ 4a &=& -2/:4 \Rightarrow a = -\frac{1}{2} \end{eqnarray} The quadratic function can be written as: $$ f(x) = -\frac{1}{2}x^2 + x + \frac{3}{2}$$ - Example 5 Determine the quadratic function which is defined with three points \(A(0,1), B(-1,2)\), and \(C(2,17)\).
Solution The solution of this exercise can be found by inserting the coordinated of defined points into the general quadratic function \(f(x) = ax^2 + bx + c\). \begin{eqnarray} A(0,1) &\Rightarrow& f(0) = a\cdot 0^2 + b\cdot 0 + c = 1 \\ c &=& 1 \\ B(-1,2) &\Rightarrow& f(-1) = a\cdot (-1)^2 + b\cdot (-1) + c = 2 \\ a - b + 1 &=& 2 \Rightarrow a-b = 1\\ C(2,17) &\Rightarrow& f(2) = a\cdot 2^2 + b\cdot 2 + c = 17 \\ 4a + 2b + 1 &=& 17 \Rightarrow 4a +2b = 16/:2 \\ 2a+b &=& 8 \end{eqnarray} The problem is reduced to solving the three equations with three unkowns. However, we have already found the value of \(c\) which is equal to 1. So, our problem is now reduced to two equations with two unkwons. \begin{eqnarray} a - b &=& 1 \Rightarrow a = 1+b\\ 2a + b &=& 8 \\ 2(1+b)+ b &=& 8\\ 2+2b + b &=& 8 \\ 3b &=& 6/:3\\ b &=& 2\\ a &=& 1 + 2 = 3\\ \end{eqnarray} Now that coefficients \(a,b\) and \(c\) are determined the quadratic function can be written as: $$ f(x) = 3x^2 +2x + 1$$ - Example 6 Write the quadratic function if the roots of this functions are 2 and -3 while the value of the leading coefficient \(a\) is 5.
Solution The quadratic function can be written as: $$ f(x) = a(x-x_1)(x-x_2)$$ where \(x_1\) and \(x_2\) are roots of quadratic function and \(a\) is the leading coefficient. By inserting values of roots and leading coefficient the quadratic function can be written as: $$ f(x) = 5(x-2)(x+3)$$ This function can also be written in traditional form: \begin{eqnarray} f(x) &=& 5(x^2 + 3x -2x-6) \\ f(x) &=& 5(x^2 + x -6)\\ f(x) &=& 5x^2 + 5x - 30 \end{eqnarray} - Example 7 Analyze the flow of quadratic function \(f(x) = \frac{3}{2}(x-4)(x+2)\)
Solution Since the quadratic function is written in this form we can immediately extract the useful infromation i.e. leading coefficient \(a =\frac{3}{2}\) and the roots \(x_1= 4\) and \(x_2 = -2\). However to find the vertex of the quadratic function the function must be rewritten into traditional form. \begin{eqnarray} f(x) &=& \frac{3}{2}(x^2 + 2x - 4x - 8)\\ f(x) &=& \frac{3}{2}(x^2 - 2x -8)\\ f(x) &=& \frac{3}{2}x^2 -3x - 12 \end{eqnarray} Now we have determined the coefficients \(a, b\) and \(c\) and with them the coordinates of the vertex can be determined. \begin{eqnarray} a &=& \frac{3}{2}, b= -3, c = -12 \\ &&T \left(-\frac{b}{2a}, \frac{4ac - b^2}{4a}\right)\\ -\frac{b}{2a} &=& -\frac{-3}{2\frac{3}{2}} = \frac{3}{3} = 1\\ \frac{4ac - b^2}{4a} &=& \frac{4\frac{3}{2}\cdot (-12) - (-3)^2}{4\frac{3}{2}}\\ \frac{4ac - b^2}{4a} &=& \frac{-72-9}{6} = -\frac{81}{6} = -\frac{27}{2}\\ &&T\left(1, -\frac{27}{2}\right) \end{eqnarray} The flow of the quadratic function is shown in the following table. The general look of the function is obtained by observing the quadratic function itself and the coordinates of the vertex. The coefficient \(a\) is greater than 0 which means that the function is concave up parabola. The shape of parabola suggests that the vertex point of the parabola is the minimum value. Looking from the perspective of \(x-axis\) as the \(x\) value icreases from \(-\infty\) to \(x_0\) the function value decereases. At the \(x_0\) i.e. vertex point the function has the minimum value and as the \(x\) value increases from \(x_0\) to \(+\infty\) the function value also increases.\(x\) \(-\infty\) \(\nearrow\) \(x_0 = 1\) \(\nearrow\) \(+\infty\) \(x\) \(+\infty\) \(\searrow\) \(y_0 = -\frac{27}{2}\) \(\nearrow\) \(+\infty\) - Example 8 Determine the mutual position of line \(f(x) = -x + 5\) and parabola \(g(x) = x^2 -8x + 15\)
Solution Basically the intersection points of these two curves must be determined. \begin{eqnarray} x^2 - 8x + 15 &=& -x + 5 \\ x^2 - 7x + 10 &=& 0 \\ x^2 -2x -5x + 10 &=& 0 \\ x(x-2) - 5(x-2) &=& 0 \\ (x-2)(x-5) &=& 0 \\ x_1 = 2, x_2 &=& 5\\ f(2) &=& -2 + 5 = 3 \Rightarrow y_1 = 3\\ f(5) &=& -5 + 5 = 0 \Rightarrow y_2 = 0 \end{eqnarray} The line and the parabola intersect each other at points \(T_1(2,3)\) and \(T_2(5,0)\) - Example 9 The sum of two numbers is 50. Determine these two numbers so that their product is maximal.
Solution In this case those two unkonws will be labeled as \(x\) and \(y\). The problem is reduced to solve the two equations with two unkowns. \begin{eqnarray} x+y &=& 50 \Rightarrow y &=& 50-x\\ g(x) = xy = x(50-x) = -x^2 + 50x \\ &&T \left(-\frac{b}{2a}, \frac{4ac - b^2}{4a}\right)\\ -\frac{b}{2a} &=& \frac{50}{2} = 25\\ y &=& 50 -x = 50 - 25 = 25 \\ \end{eqnarray} Both unknowns are equal to 25. The maximum value of this function is at the vertex point i.e. \(g(25) &=& 25\cdot 25 = 625\) - Example 10 Determine at which points line \(y= x+1\) and parabola \(y= x^2 - 6x + 7\) intersects.
Solution \begin{eqnarray} x+1 &=& x^2 -6x + 7 \\ x^2 -7x + 6 &=& 0 \\ x^2 - x -6x + 6 &=& 0 \\ x(x-1) -6(x-6) &=& 0 \\ (x-1)(x-6)&=& 0\\ x_1 = 1, x_2 &=& 6 \\ y_1 &=& 1+1 = 2 \\ y_2 &=& 6+1 = 7 \end{eqnarray} Those two functions intersect at points \(A_1(1,2)\) and \(A_2(6,7)\) - Example 1 -
Solution -
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