We want to draw the graph of the given quadratic function. To do so, we will rewrite it in vertex form,

f (x) = a (x - h)^{2} + k,

where

a,

h,

and

k

are either positive or negative numbers.

f (x) = - 2 (x - 3)^{2} ⇕ f (x) = - 2 (x - 3)^{2} + 0

To draw the graph, we will follow four steps.

Identify the constants $a,$ $h,$ and $k .$
Plot the vertex $(h, k)$ and draw the axis of symmetry $x = h .$
Plot any point on the curve and its reflection across the axis of symmetry.
Sketch the curve.
Let's get started.

Step $1$
We will first identify the constants $a,$ $h,$ and $k .$ Recall that if $a < 0,$ the parabola will open downwards. Conversely, if $a > 0,$ the parabola will open upwards.
$Vertex Form : Function : f (x) = - a (x - h)^{2} + k f (x) = - 2 (x - 3)^{2} + 0$
We can see that $a = - 2,$ $h = 3,$ and $k = 0 .$ Since $a$ is less than $0,$ the parabola will open downwards.
Step $2$

Let's now plot the vertex $(h, k)$ and draw the axis of symmetry $x = h .$ Since we already know the values of $h$ and $k,$ we know that the vertex is $(3, 0) .$ Therefore, the axis of symmetry is the vertical line $x = 3 .$

Step $3$
We will now plot a point on the curve by choosing an $x -$ value and calculating its corresponding $y -$ value. Let's try $x = 1 .$
$f (x) = - 2 (x - 3)^{2}$
Substitute
$x = 1$
$f (1) = - 2 (1 - 3)^{2}$

▼

Simplify right-hand side

SubTerm
Subtract term
$f (1) = - 2 (- 2)^{2}$
CalcPow
Calculate power
$f (1) = - 2 (4)$
MultNegPos
$(- a) b = - a b$

$f (1) = - 8$
When $x = 1,$ we have $y = - 8 .$ Thus, the point $(1, - 8)$ lies on the curve. Let's plot this point and reflect it across the axis of symmetry.

Note that both points have the same $y -$ coordinate.

Step $4$

Finally, we will sketch the parabola which passes through the three points. Remember not to use a straightedge for this!

Step 1

Step 2

Step 3

Step 4

Step $1$

Step $2$

Step $3$

Step $4$

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