Study S3 Mathematics Quadratic Equations and Functions - Geniebook

• 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 $$x=\frac {a+b}{2}$$ $$x=a$$ $$x=a$$

### 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 Shape Coefficient of $$x^2$$ I Intercepts The points of $$x, \;y$$ coordinates T 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 $$1$$, hence it is a minimum point graph with an upward parabola x–intercept Putting $$y=0$$; we get $$x=1$$ or $$x=3$$ Line of symmetry Upon finding the mean of the x-intercepts you get a line of symmetry. $$x =\frac {1+3}{2}=2$$ y–intercept 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 $$-1$$, hence it is a maximum point graph with downward parabola x–intercept Putting $$y=0$$; we get $$x=1$$ or $$x=3$$ Line of symmetry Upon finding the mean of the x-intercepts you get a line of symmetry. \begin{align*} x =\frac {1+3}{2}=2 \end{align*} y–intercept Putting \begin{align*} x=0; \;\;y &= -(-1)(-3) =-3 \end{align*}

Example 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 $$y = (x - 2)(x + 4)$$

A)  For x-intercepts:

Upon substituting y = 0,

\begin{align*} (x - 2)(x + 4) &= 0 \\ \\ ⇒\qquad\qquad (x - 2) &= 0\qquad \text{or}\qquad(x +4) = 0 \\ \\ ⇒\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 $$(0, 8).$$

B)  Taking the mean of the x-intercepts,

\begin{align*} x &= \frac {2+(-4)}{2} \\ \\ x &= -22 \end{align*}

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

C)  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  $$(-1, -9)$$.

D)

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

## Conclusion

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