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 Graphing Quadratic Equations The general technique for graphing quadratics is the same as for graphing linear equations. However, since quadratics graph as curvy lines (called "parabolas"), rather than the straight lines generated by linear equations, there are some additional considerations: Determine whether the parabola opens upward (if a>0) or downward (if a<0). Find $\frac{-b}{2a}$, which is the x coordinate of the vertex. Find $f(\frac{-b}{2a})$, which is the y coordinate of the vertex (You could also find the y coordinate by evaluating $\frac{4ac-b^2}{4a}$). Finally find the equation of the axis of symmetry: $x=\frac{-b}{2a}$. Find the vertical intercept. Find the horizontal intercepts (if any). Plot the points you found. Plot their symmetric points and sketch the graph (Find an additional pair of symmetric points if needed). Let's use these steps in the following example: Graph f(x)=x2-4x+3 Step 1. Determine whether the graph opens up or down. Because a is positive in this quadratic, we know that the parabola will open upward. Step 2. Find the vertex and the equation for the axis of symmetry. This function is in standard form for a quadratic, so we can use the formula $x=\frac{-b}{2a}$ to find the x-value of the vertex: $x=\frac{-b}{2a}=\frac{4}{2}=2$ Now that we have the input value for the vertex, we can substitute it into the function and find the output value for the vertex: f(x)=x2-4x+3 f(2)=22-4(2)+3 f(2)=4-8+3 f(2)=-1 The vertex is (2,-1). The axis of symmetry is the vertical line through the vertex. The axis of symmetry is x=2. Step 3. Find the vertical intercept. Because this quadratic is in standard form, we know that the vertical intercept has an output value equal to a constant value, c. In this case, the vertical intercept is (0,3) Step 4. Find the horizontal intercepts (if any) Horizontal intercepts will happen when the output value is equal to zero. Substitute zero for the output variable and solve: f(x)=x2-4x+3 0=x2-4x+3 a=1, b=-4, c=3 Use the quadratic formula $x=\frac{-b\;\pm\;\sqrt{b^2-4ac}}{2a}$ $x=\frac{4\;\pm\;\sqrt{4^2-4(1)(3)}}{2(1)}$ $x=\frac{4\;\pm\;\sqrt{16-12}}{2}$ $x=\frac{4\;\pm\;\sqrt{4}}{2}$ $x=\frac{4\;+\;2}{2}=\frac{6}{2}=3$ and $x=\frac{4\;-\;2}{2}=\frac{2}{2}=1$ The horizontal intercepts are (3,0) and (1,0) Step 5. Plot the points you found.