# quadratic function and vertex help!

• Mar 21st 2012, 09:55 PM
noork85
quadratic function and vertex help!
ok this is a very confusing topic. im doing the review for my midterm and encountered a problem, heres the question:

g(x)=x^2-2x+5 professors answer =vertex is (1,4) and x-intercept is none and y--intercept is (0,5)

the only thing i got right was the y-intercept. for the vertex im getting (1,6), and why is there not an x-intercept?

id appreciate any help? thank u!
• Mar 21st 2012, 10:09 PM
beebe
Re: quadratic function and vertex help!
You got the x part of the vertex correct. To find the y coordinate, plug x=1 back in. What's g(1)?

You find the x-intercept by setting g(x)=0 and solving. Can you solve it? What happens when you use the quadratic formula to find the x-intercepts. Also, you may notice that the coefficient on the $x^2$ is positive, which means the parabola opens upwards, and the vertex is above the x-axis (4 is a positive number). Since the vertex is the lowest point of the parabola and it only goes up from there, it can never cross the x-axis.
• Mar 21st 2012, 10:09 PM
princeps
Re: quadratic function and vertex help!
Quote:

Originally Posted by noork85
ok this is a very confusing topic. im doing the review for my midterm and encountered a problem, heres the question:

g(x)=x^2-2x+5 professors answer =vertex is (1,4) and x-intercept is none and y--intercept is (0,5)

the only thing i got right was the y-intercept. for the vertex im getting (1,6), and why is there not an x-intercept?

id appreciate any help? thank u!

$g(x)=ax^2+bx+c$

hence :

$a=1 , ~ b=-2, ~c=5$

therefore :

$D=b^2-4ac=-16$

for vertex we have :

$V\left(\frac{-b}{2a},\frac{-D}{4a}\right) \Rightarrow V(1,4)$

Since $D < 0$ graph of the function and x-axis have no common points .
• Mar 21st 2012, 10:12 PM
noork85
Re: quadratic function and vertex help!
can i use this formula for all that are in the format of g(x)=ax^2+bx+c?

Quote:

Originally Posted by princeps
$g(x)=ax^2+bx+c$

hence :

$a=1 , ~ b=-2, ~c=5$

therefore :

$D=b^2-4ac=-16$

for vertex we have :

$V\left(\frac{-b}{2a},\frac{-D}{4a}\right) \Rightarrow V(1,4)$

Since $D < 0$ graph of the function and x-axis have no common points .

• Mar 21st 2012, 10:13 PM
noork85
Re: quadratic function and vertex help!

and whats the D stand for? i know is vertex?

still lost, where does the -16 come from? can u break down step by tiny step?
• Mar 21st 2012, 10:31 PM
beebe
Re: quadratic function and vertex help!
D is the discriminant. If you haven't learned it yet, when D>0, then there are two roots. When D=0, there is one root. When D<0, then there are no roots.

Notice that the discriminant is the part inside of the square root for the quadratic formula. $x=\frac{-b \pm \sqrt {b^2-4ac}}{2a}$ As princeps pointed out, $b^2-4ac$ is a negative number. You can't take a square root of a negative number and get a real number for a result. In other words, you can't square any real number and get -16.

Step by step:
$D=b^2-4ac$
You know what a, b, and c are, so just substitute and do the arithmetic.
$D=(-2)^2-4*1*5=4-20=-16$

Alternately, if you had gone straight to the quadratic formula to find the x-intercepts, you would get:
$x=\frac{-b \pm \sqrt {b^2-4ac}}{2a}$
$x=\frac{-(-2) \pm \sqrt {(-2)^2-4*1*5}}{2*1}$
$x=\frac{2 \pm \sqrt{-16}}{2}$
The quadratic formula will always give you the x-intercepts for a quadratic function (if they exist), but you can't find the square root of -16, so there are no real number x-intercepts.
• Mar 21st 2012, 10:52 PM
noork85
Re: quadratic function and vertex help!
oh thank u so much! that really cleared it up. i didnt learn anything about discriminitives. but yes, i shouldve just gone straight to the quadratic formula to solve it.
• Mar 21st 2012, 10:57 PM
noork85
Re: quadratic function and vertex help!
can u also please explain the vertex part.
• Mar 23rd 2012, 10:40 AM
beebe
Re: quadratic function and vertex help!
It might be helpful if you showed how you got to your answer.

You found the x value for the vertex already. x=1 is correct. Let's split the denominator of the quadratic formula so that it's $x=\frac{-b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}$. Notice that you add or subtract $\frac{\sqrt{b^2-4ac}}{2a}$ from $\frac{-b}{2a}$. Also remember that the vertex of a parabola is on it's axis of symmetry. So if you add or subtract the same amount from a number to get the x-intercepts, then the axis of symmetry is at $x=\frac{-b}{2a}$.

Plug in the numbers and do the arithmetic and you get x=1, which you already knew. At this point, noork85 went a seperate route to find the y value, which I'm not familiar with, so here's the way that I do it, which seems more intuitive:

Where is the parabola at when x=1? It's at g(1). All you have to do is go back to your original function and plug 1 in for x and do the arithmetic.
• Mar 27th 2012, 02:27 PM
noork85
Re: quadratic function and vertex help!
thank u so very much! very detailed and helpful!