1. ## graphing a function

can someone show me how i go about graphing this?

f(x) = 15 - x - x^2

2. Originally Posted by el123
can someone show me how i go about graphing this?

f(x) = 15 - x - x^2

3. thanks for that link , so i've just attempted a graph looks like a squiggly line lol.

I was just curious, i can write f(x) = 15 - x - x^2
or y = 15 - x - x^2

Is that ok then i just make a table with my x values and substitue them ionto the equation to find each value of y for any given x value?

4. Originally Posted by el123
thanks for that link , so i've just attempted a graph looks like a squiggly line lol.

I was just curious, i can write f(x) = 15 - x - x^2
or y = 15 - x - x^2

Is that ok then i just make a table with my x values and substitue them ionto the equation to find each value of y for any given x value?
That should be fine, yes.

5. Originally Posted by el123
can someone show me how i go about graphing this?

f(x) = 15 - x - x^2
If your graph looks like a "squiggly line," then you got your signs mixed up somewhere.

To graph this, you could rewrite in vertex form:
$f(x) = a(x - h)^2 + k$

So,
$f(x) = 15 - x - x^2$
$f(x) = -x^2 - x + 15$
$f(x) = -(x^2 + x) + 15$
$f(x) = -\left(x^2 + x + \frac{1}{4}\right) + 15 + \frac{1}{4}$
$f(x) = -\left(x + \frac{1}{2}\right)^2 + \frac{61}{4}$

Since the leading coefficient was negative, the parabola opens downward. The vertex is given as (h, k), so it's $\left(-\frac{1}{2},\;\frac{61}{4}\right)$. You can also find the x- and y- intercepts to get additional points.

01

6. Originally Posted by yeongil
If your graph looks like a "squiggly line," then you got your signs mixed up somewhere.

To graph this, you could rewrite in vertex form:
$f(x) = a(x - h)^2 + k$

So,
$f(x) = 15 - x - x^2$
$f(x) = -x^2 - x + 15$
$f(x) = -(x^2 + x) + 15$
$f(x) = -\left(x^2 + x + \frac{1}{4}\right) + 15 + \frac{1}{4}$
$f(x) = -\left(x + \frac{1}{2}\right)^2 + \frac{61}{4}$

Since the leading coefficient was negative, the parabola opens downward. The vertex is given as (h, k), so it's $\left(-\frac{1}{2},\;\frac{61}{4}\right)$. You can also find the x- and y- intercepts to get additional points.

01
Hello : look here the graph Thank you