• Determining the shape of a Quadratic Graph
• Finding the intercepts of a Quadratic Graph
• Finding the line of symmetry and the turning point of a Quadratic Graph
• Sketching Graphs of Quadratic Functions

### Determining the shape of a Quadratic Graph

Ideally, the name of the shape of the Quadratic graph is called a “Parabola graph”.

$y=ax^2+bx+c$

The given graph above is a minimum point graph.

If the value of $a>0$, the graph has a minimum point.

$y=ax^2+bx+c$

The given graph above is a maximum point graph.

If the value of $a<0$, the graph has a maximum point.

### Finding the y-Intercept of a Quadratic Graph

A quadratic graph will always have only one y-intercept.

To find the y-intercept, substitute $x = 0$ and solve for $y$.

$y=ax^2+bx+c$

Putting in $x = 0$

We can find the values of the points where the graph cuts the y-axis.

### Finding the y-Intercept of a Quadratic Graph

A quadratic graph may have $2, 1 \;or\; 0$ x-intercepts.

In the first case, as seen, there are two points where the graph cuts off the x-axis.

In the second case, as seen, there is one point where the graph cuts off the x-axis.

In the third case, as seen, there is no point where the graph cuts off the x-axis.

To find the x-intercept, substitute $y = 0$ and solve for $x$.

$y=ax^2+bx+c$

Putting in $y = 0$,

We can find the values of the points where the graph cuts the x-axis.

### Finding the line of symmetry of a Quadratic Graph

The line of symmetry is the line that passes through $x$ and splits the entire graph into two mirror halves.

Under this, there are 3 different cases:

Using this simple table below we can find the equation for the line of symmetry

2  x-intercepts 1  x-intercept No  x-intercepts
$\displaystyle{x=\frac {a+b}{2}}$ $\displaystyle{x=a}$ $\displaystyle{x=a}$

### Finding the turning point of a Quadratic Graph

A turning point is a point on the graph where it changes its direction. It can either be the turning at the highest point or the lowest point.

For every turning point, the x-coordinate always lies on the line of symmetry.

In the case of a linear equation graph, we need to know only the x-intercept and the y-intercept to sketch the graph.

However, in the case of Quadratic graphs, you need a few more parameters to sketch the graph.

For a graph given by the formula $y=ax^2+bx+c$,

Remember the Acronym S I T.

S I Shape Coefficient of $x^2$ Intercepts The points of $x, \;y$ coordinates. Turning Point The maximum or minimum turning point

### Graphs of the form y = (x – h)(x – k)

Example: For the graph of $y= (x-1)(x-3)$ shown above,

Shape Since the coefficient of $x^2$ is $1$, it is a minimum point graph with an upward parabola Putting $y=0$; we get $x=1$ or $x=3$. Upon finding the mean of the x-intercepts you get a line of symmetry.  $\displaystyle{x =\frac {1+3}{2}=2}$ Putting \begin{align*} x=0; \;\;y &= (-1)(-3) =3 \end{align*}

### Graphs of the form –y = (x – h)(x – k)

Example: For the graph of $y= -(x-1)(x-3)$ shown above,

Shape Since the coefficient of $x^2$ is $-1$, it is a maximum point graph with a downward parabola. Putting $y=0$; we get $x=1$ or $x=3$. Upon finding the mean of the x-intercepts you get a line of symmetry.  $\displaystyle{x =\frac {1+3}{2}=2}$ Putting \begin{align*} x=0; \;\;y &= -(-1)(-3) =-3 \end{align*}

Question 1:

Given the quadratic function $y = (x - 2)(x + 4)$

1. Find the coordinates of the x-intercepts and y-intercepts.
2. State the equation of the line of symmetry of the graph.
3. Find the coordinates of the turning point of the graph. State whether it is a maximum or minimum.
4. Sketch the graph.

Solution:

The given equation is $y = (x - 2)(x + 4)$

1. For x-intercepts:

Upon substituting y = 0,

\begin{align*} (x - 2)(x + 4) &= 0 \\ \\ \implies\qquad\qquad (x - 2) &= 0\qquad \text{or}\qquad(x +4) = 0 \\ \\ \implies\qquad\qquad\qquad\;\; x &= 2\qquad \text{or}\qquad x = -4 \end{align*}

Therefore, there are two x-intercepts obtained: $(2, 0)$ & $(-4, 0)$

For y-intercepts:

Upon substituting $x = 0$,

\begin{align*} y &= (-2)(4) \\ \\ y &= -8 \end{align*}

Therefore, the y-intercept is $(0, 8).$

1. Taking the mean of the x-intercepts,

\begin{align*} x &= \frac {2+(-4)}{2} \\ \\ x &= {-2\over2} \end{align*}

∴  \begin{align*} x= -1 \end{align*} is the equation of the line of symmetry of the graph.

1. Since for the given equation, the coefficient of $x^2 > 0$, the graph is an upward opening parabola and has a minimum turning point.

Putting $x = -1$ in the original equation,

\begin{align*} y &= (-1 - 2)(-1 + 4)\\ \\ y &= (-3)(3) \\ \\ y &= -9 \end{align*}

Hence the turning point is at  $(-1, -9)$.

1.
S Upward opening parabola x-intercepts $(2, 0 )$ & $(-4, 0 )$ y-intercept $(0,8)$ $(-1, -9 )$

## Conclusion

Continue Learning
Further Trigonometry Quadratic Equations And Functions
Linear Inequalities Laws of Indices
Coordinate Geometry Graphs Of Functions And Graphical Solution
Applications Of Trigonometry
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