To graph the quadratic function you have to determine the coordinates of its vertex and the roots (x-intercepts)
1) To find the roots of the function, the first step is to zero it:
[tex]\begin{gathered} 3x^2+5x-2-15=15-15 \\ 3x^2+5x-17=0 \end{gathered}[/tex]Next, you have to identify the coefficients of the function:
"a" represents the coefficient of the squared term, for this function a= 3
"b" is the coefficient of the x-term, for this function b= 5
"c" is the constant of this function, in this case, c= -17
Use the quadratic function to calculate the possible x-intercepts:
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]Replace the formula with the coefficients:
[tex]\begin{gathered} x=\frac{-5\pm\sqrt[]{5^2-4\cdot3\cdot(-17)}}{2\cdot3} \\ x=\frac{-5\pm\sqrt[]{25+204}}{6} \\ x=\frac{-5\pm\sqrt[]{229}}{6} \end{gathered}[/tex]Solve the addition and subtraction separately:
Addition:
[tex]\begin{gathered} x=\frac{-5+\sqrt[]{229}}{6} \\ x=1.689 \end{gathered}[/tex]Subtraction:
[tex]\begin{gathered} x=\frac{-5-\sqrt[]{229}}{6} \\ x=-3.355 \end{gathered}[/tex]The roots of the function are x=1.689 and x=-3.355
2) To determine the coordinates of the vertex, you have to use the following formula to determine the x-coordinate (h):
[tex]h=\frac{-b}{2a}[/tex]Replace it with b=5 and a=3
[tex]h=-\frac{5}{2\cdot3}=-\frac{5}{6}[/tex]Next, replace the value of x into the function to calculate the corresponding y-coordinate (k)
[tex]\begin{gathered} k=3x^2+5x-17 \\ k=3(-\frac{5}{6})^2+5\cdot(-\frac{5}{6})-17 \\ k=3\cdot\frac{25}{36}-\frac{25}{6}-17 \\ k=\frac{25}{12}-\frac{25}{6}-17 \\ k=-\frac{229}{12}\approx-19.083 \end{gathered}[/tex]The coordinates of the vertex are (-0.833,-19.083)
Plot the points and graph the quadratic function:
Note that the coefficient of the quadratic term is positive, this indicates that the parabola opens upwards.
