# Reading: Graphs of Quadratics

Just as we drew pictures of the solutions for lines or linear equations, we can draw a picture of solution to quadratics as well. One way we can do that is to make a table of values.

We will test 5 values to get an idea of shape
y = (0)

Plot points on graph: (0, 3), (1, 0), (2, − 1), (3, 0), and (4, 3). Connect the dots with a smooth curve.
Our Solution
When we have

The above method to graph a parabola works for any equation, however, it can be very tedious to find all the correct points to get the correct bend and shape. For this reason we identify several key points on a graph and in the equation to help us graph parabolas more efficiently. These key points are described below.

Graph the

Graph the

Tyler Wallace, Beginning and Intermediate Algebra, " Graphs of Quadratics," licensed under a CC-BY license.

### Example 1

y = x^{2}− 4x +3 Make a table of valuesx |
y |

0 | |

1 | |

2 | |

3 | |

4 |

^{2}+4(0)+ 3= 0 − 0+ 3=3 Plug 0 in for x and evaluate y =(1)^{2}− 4(1)+ 3= 1 − 4+ 3=0 Plug 1 in for x and evaluate y = (2)^{2}− 4(2)+ 3=4 − 8+ 3= − 1 Plug 2 in for x and evaluate y =(3)^{2}− 4(3)+ 3= 9 − 12 + 3=0 Plug 3 in for x and evaluate y =(4)^{2}− 4(4)+ 3= 16 − 16 + 3=3 Plug 4 in for x and evaluate Our completed table.x |
y |

0 | 3 |

1 | 0 |

2 | –1 |

3 | 0 |

4 | 3 |

*x*^{2}in our equations, the graph will no longer be a straight line. Quadratics have a graph that looks like a U shape that is called a parabola.### World View Note

The first major female mathematician was Hypatia of Egypt who was born around AD 370. She studied conic sections. The parabola is one type of conic section.The above method to graph a parabola works for any equation, however, it can be very tedious to find all the correct points to get the correct bend and shape. For this reason we identify several key points on a graph and in the equation to help us graph parabolas more efficiently. These key points are described below.

**Point A:***y*-intercept: Where the graph crosses the vertical*y*-axis.**Points B and C:***x*-intercepts: Where the graph crosses the horizontal*x*-axis**Point D:**Vertex: The point where the graph curves and changes directions. We will use the following method to find each of the points on our parabola. To graph the equation*y*=*ax*^{2}+*bx*+*c*, find the following points*y*-intercept: Found by making*x*= 0,this simplifies down to*y*=*c**x*-intercepts: Found by making*y*= 0, this means solving 0=*ax*^{2}+*bx*+*c*- vertex: Let [latex]\displaystyle{x}=\frac{{-{b}}}{{{2}{a}}}[/latex] to find
*x*. Then plug this value into the equation to find*y*.

### Example 2

Find the key points | [latex]\displaystyle{y}={x}^{{2}}+{4}{x}+{3}[/latex] | |

y = c is the y-intercept |
[latex]\displaystyle{y}={3}[/latex] | |

To find the x-intercept we solve the equation |
[latex]\displaystyle{0}={x}^{{2}}+{4}{x}+{3}[/latex] | |

Factor | [latex]\displaystyle{0}={({x}+{3})}{({x}+{1})}[/latex] | |

Set each factor equal to zero | [latex]\displaystyle{x}+{3}={0}[/latex] | [latex]\displaystyle{x}+{1}={0}[/latex] |

Solve each equation | [latex]\displaystyle{x}+{3}-{3}={0}-{3}[/latex] | [latex]\displaystyle{x}+{1}-{1}={0}-{1}[/latex] |

Our x-intercepts |
[latex]\displaystyle{x}=-{3}[/latex] | [latex]\displaystyle{x}=-{1}[/latex] |

To find the vertex, first use [latex]\displaystyle{x}=\frac{{-{b}}}{{{2}{a}}}[/latex] | [latex]\displaystyle{x}=\frac{{-{4}}}{{{2}{({1})}}}=\frac{{-{4}}}{{2}}=-{2}[/latex] | |

Plug this answer into the equation to find the y-coordinate |
[latex]\displaystyle{y}={(-{2})}^{{2}}+{4}{(-{2})}+{3}[/latex] | |

Evaluate | [latex]\displaystyle{y}={4}-{8}+{3}[/latex] | |

y-value of the vertex |
[latex]\displaystyle{y}=-{1}[/latex] | |

Vertex as a point | [latex]\displaystyle{(-{2},-{1})}[/latex] |

*y*-intercept at 3, the*x*-intercepts at −3 and −1, and the vertex at ( −2, −1). Connect the dots with a smooth curve in a U shape to get our parabola. Our Solution If the*a*in*y*=*ax*^{2}+*bx*+*c*is a negative value, the parabola will end up being an upside-down U. The process to graph it is identical, we just need to be very careful of how our signs operate. Remember, if a is negative, then*ax*^{2}will also be negative because we only square the*x*, not the*a*.### Example 3

Find key points | [latex]\displaystyle{y}=-{3}{x}^{{2}}+{12}{x}-{9}[/latex] | |

y-intercept is y = c |
[latex]\displaystyle{y}=-{9}[/latex] | |

To find x-intercept solve this equation |
[latex]\displaystyle{0}=-{3}{x}^{{2}}+{12}{x}-{9}[/latex] | |

Factor out GCF first, then factor the rest | [latex]\displaystyle{0}=-{3}{({x}^{{2}}-{4}{x}+{3})}[/latex] | |

[latex]\displaystyle{0}=-{3}{({x}-{3})}{({x}-{1})}[/latex] | ||

Set each factor with a variable equal to zero | [latex]\displaystyle{x}-{3}={0}[/latex] | [latex]\displaystyle{x}-{1}={0}[/latex] |

Solve each equation | [latex]\displaystyle{x}-{3}+{3}={0}+{3}[/latex] | [latex]\displaystyle{x}-{1}+{1}={0}+{1}[/latex] |

Our x-intercepts |
[latex]\displaystyle{x}={3}[/latex] | [latex]\displaystyle{x}={1}[/latex] |

To find the vertex, first use [latex]\displaystyle{x}=\frac{{-{b}}}{{{2}{a}}}[/latex] | [latex]\displaystyle{x}=\frac{{-{12}}}{{{2}{(-{3})}}}=\frac{{-{12}}}{{-{6}}}={2}[/latex] | |

Plug this value into the equation to find the y-coordinate |
[latex]\displaystyle{y}=-{3}{({2})}^{{2}}+{12}{({2})}-{9}[/latex] | |

Evaluate | [latex]\displaystyle{y}=-{3}{({4})}+{24}-{9}[/latex] | |

[latex]\displaystyle{y}=-{12}+{24}-{9}[/latex] | ||

y-value of the vertex |
[latex]\displaystyle{y}={3}[/latex] | |

Vertex as a point | [latex]\displaystyle{({2},{3})}[/latex] |

*y*-intercept at –9, the*x*-intercepts at 3 and 1, and the vertex at (2, 3). Connect the dots with smooth curve in an upside-down U shape to get our parabola. Our SolutionTyler Wallace, Beginning and Intermediate Algebra, " Graphs of Quadratics," licensed under a CC-BY license.