# Thread: Functions and finding x-intercepts and vertex.

1. ## Functions and finding x-intercepts and vertex.

Hi, I'm new to Math Help Forum and was wondering if you could help me with functions and finding the x-intercepts and the coordinates of the vertex.

I'm having a hard time with these questions:

y=x^2+7x

and

y= 15x^2+16-7

Can you please explain step by step. I use the quadratic formula but my anwser seems be be wrong.

thanks for taking time to read this, hopefully you can help.

2. Originally Posted by kimmybimmy
Hi, I'm new to Math Help Forum and was wondering if you could help me with functions and finding the x-intercepts <<<<<< y = 0
and the coordinates of the vertex. <<<<<<<< complete the square

I'm having a hard time with these questions:

y=x^2+7x

and

y= 15x^2+16-7

Can you please explain step by step. I use the quadratic formula but my anwser seems be be wrong.

thanks for taking time to read this, hopefully you can help.
I'll take the 2nd example, then the 1st example is left for you:

1. I assume that you mean:

$y = 15x^2+16x-7$

2. x-intercept: y = 0

$0=15x^2+16x-7$ Use the quadratic formula

$x = \frac{-16 \pm \sqrt{(-16)^2-4 \cdot 15 \cdot (-7)}}{2 \cdot 15}$ Simplify!

I've got $x = \frac13~\vee~x=-\frac75$

3. Coordinates of the vertex:

$y=15x^2+16x-7 = 15\left(x^2+\frac{16}{15}x\bold{\color{red}+\left( \frac8{15}\right)^2-\left(\frac8{15}\right)^2}\right) - 7$

$y=15x^2+16x-7 = 15\left(\left(x+\bold{\color{red}\left(\frac8{15}\ right)}\right)^2 \bold{\color{red}-\left(\frac8{15}\right)^2}\right) - 7$

$y=15x^2+16x-7 = 15\left(x+\bold{\color{red}\left(\frac8{15}\right) }\right)^2 \bold{\color{red}-\left(\frac{64}{15}\right)} - 7$

$y=15x^2+16x-7 = 15\left(x+\left(\frac8{15}\right)\right)^2 -\frac{169}{15}$

Therefore the vertex is at $V\left(-\frac8{15}~,~-\frac{169}{15}\right)$

4. And now it's your turn!

3. Since you going to have to complete the square to find the vertex anyway, I would prefer using that to find the y- intercepts rather than the quadratic formula.

$y= x^2+ 7x= 0$

Half of 7 is 7/2 and the square of that is 49/4:

$y= x^2+ 7x+ \frac{49}{4}- \frac{49}{4}= 0$

$y= (x+ \frac{7}{2})^2- \frac{49}{4}= 0$

At this point you can argue that $(x+ \frac{7}{2})^2$ is never negative so y is always $-\frac{49}{4}$ "plus something". That tells you than the lowest value for y is $-49/4$ which occurs when $x+ \frac{7}{2}= 0$ or $x= -\frac{7}{2}$.

The vertex is at $\left(-\frac{7}{2}, -\frac{49}{4}\right)$

Now, going back to $(x+ \frac{7}{2})^2- \frac{49}{4}= 0$,
adding $\frac{49}{4}$ to each side gives
$(x+ \frac{7}{2})^2= \frac{49}{4}$
and so

$x+ \frac{7}{2}= \pm\frac{7}{2}$

so that $x= -\frac{7}{2}+ \frac{7}{2}= 0$
or $x= -\frac{7}{2}- \frac{7}{2}= -7$.

Of course, for this problem, the simplest way to do it is to factor:
$x^2+ 7x= x(x+ 7)= 0$ so either x= 0 or x+ 7= 0 and x= -7.