NZ Level 7 (NZC) Level 2 (NCEA)
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Quadratic graphing
Lesson

Let's review what we know about graphing quadratic functions. Quadratic functions can be written in each of the following forms:

  • $y=ax^2+bx+c$y=ax2+bx+c - This is known as the general form or $y$y-intercept form.
  • $y=a\left(x-d\right)\left(x-e\right)$y=a(xd)(xe) - This is known as the factored or $x$x-intercept form.
  • $y=a\left(x-h\right)^2+k$y=a(xh)2+k - This is known as the turning point form.

Each of these names gives us a clue to the various features we can identify on a parabola.

When graphing a quadratic, no matter in which form it is presented, we must find the $x$x-intercepts, the $y$y-intercept, and the turning point.

 

Graphing in the Form $y=a\left(x-d\right)\left(x-e\right)$y=a(xd)(xe)

Example 1

Consider the parabola $y=\left(x+1\right)\left(x-3\right)$y=(x+1)(x3).

  1. Find the $y$y value of the $y$y-intercept.

  2. Find the $x$x values of the $x$x-intercepts.

    Write all solutions on the same line separated by a comma.

  3. State the equation of the axis of symmetry.

  4. Find the coordinates of the vertex.

    Vertex $=$=$\left(\editable{},\editable{}\right)$(,)

  5. Graph the parabola.

    Loading Graph...

 

Graphing in the Form $y=ax^2+bx+c$y=ax2+bx+c

Example 2

Consider the parabola $y=x^2-6x+8$y=x26x+8.

a. Determine the value of the $y$y-intercept.

The $y$y-intercept is located along the $y$y-axis, where $x=0$x=0.

Substituting $x=0$x=0 into our function gives us $y=0^2-6\times0+8$y=026×0+8.

The $y$y-intercept is $8$8.

b. Determine the coordinates of the $x$x-intercepts.

It is easiest to find the $x$x-intercepts by first factorising our quadratic function. If we cannot factorise, we might like to use the quadratic formula instead. 

Factorising gives us $y=\left(x-4\right)\left(x-2\right)$y=(x4)(x2).

And as we saw in the first example, this gives us $x$x-intercepts of $(4,0)$(4,0) and $(2,0)$(2,0).

c. Determine the coordinates of the turning point.

We begin by finding the $x$x value of the turning point and there are a number of ways to do this. Either we can use the vertex formula or we can simply find the midpoint between the two $x$x-intercepts.

Halfway between $(4,0)$(4,0) and $(2,0)$(2,0) is $x=3$x=3.

To the find the $y$y value, we substitute $x=3$x=3 back into our equation (and we're welcome to use the form given to us or the factorised version we created in part (b)).

$y=3^2-6\times3+8$y=326×3+8

$y=-1$y=1

Now that we have all three features, we can plot the points on a graph and join them with a smooth curve.

 

Graphing in the Form $y=a\left(x-h\right)^2+k$y=a(xh)2+k

Example 3

Consider the parabola $y=-\left(x+1\right)^2+4$y=(x+1)2+4.

a. Determine the value of the $y$y-intercept.

Once again, we find the $y$y-intercept by substituting $x=0$x=0 into our function.

$y=-\left(0+1\right)^2+4$y=(0+1)2+4

$y=3$y=3

b. Determine the coordinates of the $x$x-intercepts.

The easiest way to do this is to solve $0=-\left(x+1\right)^2+4$0=(x+1)2+4.

$0$0 $=$= $-\left(x+1\right)^2+4$(x+1)2+4
$-4$4 $=$= $-\left(x+1\right)^2$(x+1)2
$4$4 $=$= $\left(x+1\right)^2$(x+1)2
$x+1$x+1 $=$= $\pm2$±2
$x$x $=$= $1$1
$x$x $=$=

$-3$3

So the $x$x-intercepts are $1,0$1,0 and $-3,0$3,0.

c. Determine the coordinates of the turning point.

Since the function is already in turning point form, we simply read it from the equation and we get $-1,4$1,4.

Now that we have all three features, we can plot the points on a graph and join them with a smooth curve.

 

Example 4

Consider the parabola $y=\left(x+1\right)\left(x-3\right)$y=(x+1)(x3).

  1. Find the $y$y value of the $y$y-intercept.

  2. Find the $x$x values of the $x$x-intercepts.

    Write all solutions on the same line separated by a comma.

  3. State the equation of the axis of symmetry.

  4. Find the coordinates of the vertex.

    Vertex $=$=$\left(\editable{},\editable{}\right)$(,)

  5. Graph the parabola.

    Loading Graph...

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