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## Algebra 1

### Unit 14: Lesson 11

Features & forms of quadratic functions- Forms & features of quadratic functions
- Worked examples: Forms & features of quadratic functions
- Features of quadratic functions: strategy
- Vertex & axis of symmetry of a parabola
- Finding features of quadratic functions
- Features of quadratic functions
- Graph parabolas in all forms
- Interpret quadratic models: Factored form
- Interpret quadratic models: Vertex form
- Interpret quadratic models
- Graphing quadratics review
- Creativity break: How does creativity play a role in your everyday life?

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# Finding features of quadratic functions

Sal finds the zeros, the vertex, & the line of symmetry of quadratic functions given in vertex form, factored form, & standard form.

## Want to join the conversation?

- if the quadratic function is a negative wouldn't the loop face down(13 votes)
- If the coefficient of the squared term in the quadratic function is negative, then your parabola will face down and vice verse.(14 votes)

- I've been searching for over half an hour and I still can't figure this out. In the function form f(x)=a(x-p)^2+q, how do I solve for A using another point on the parabola? I know that (p,q) is the vertex, and x=p, which is the axis of symmetry, but I still can't solve for an a value. I'm in 20-1, btw.(12 votes)
- How do you figure out the x functions when trying to find the y values of a parabola?(7 votes)
- It depends on what you are trying to find out. If you want to find out the zeros, then you substitute 0 for y and solve for x by converting it into factored form. You have to convert the function into either standard, vertex, or factored form depending on what you want to find out. If you still don't understand what I am saying, then you can ask me to rephrase this paragraph into a different terminology.(10 votes)

- For another scenario how would we find the vertex for standard form. It wasn't explained in this video and i think it should've(4 votes)
- x value is -b/2a (think of the first part of quadratic formula), find this and substitute the value into equation to find y, thus you have the vertex.

Or you could complete the square and get into vertex form and just read it.(2 votes)

- y=-(x-2)²+3

please explain

$$$(2 votes)- Quadratic equations are written in vertex form as:

y=a(x-h)^2+k

where (h,k) represent the vertex of the parabola, and the sign of a represents if the graph of parabola is open upwards or downwards.

In your equation y = -(x-2)^2+3,

Vertex(h,k)= (2,-3)

Since a=-1, this tells us that the graph will be open downwards.(5 votes)

- For the question in the format of 'vertex form,' how do we get the zeros?(2 votes)
- Once you have it in vertex form you should have something like (x - h)^2 + k = 0 (since zeros are where f(x) = 0), so you solve from farthest from x to closest, so subtract k, (x-h)^2 = -k, take square root, so x - h = ± √-k, and finally add h, so x = h ± √-k. The hope is that k will usually be negative on the left and positive on the right to get solutions.(4 votes)

- Are the zeros the x intercepts?(2 votes)
- Yes, provided the zeros are real numbers, then they are the x-intercepts for the parabola.(4 votes)

- How to write a functions that represents a graph(2 votes)
- you find the y intercept then substitute that for c then use 2 points to find a and b(1 vote)

- could we find the vertex if the yellow equation wasnt intercepting the x axis at all?(1 vote)
- I guess if the equation was, for example, f(t) = (t-5)^2 +4, this is the vertex form of the equation so vertex is (5,4).

Explanation: We need to find the lowest point possible on the y-axis. It is at the lowest when f(t) is the lowest possible number, that's when (t-5)^2 = 0 meaning when t = 5.

Hope that helps?(2 votes)

- In one of the exercises I got:

f(t) = -(t - 2)(t - 15)

And I was supposed to find the vertex coordinates.

I've come to the solution: (17/2, -169/4)

However the correct answer is (17/2, 169/4) -> "y" value is positive instead of negative.

I can't find where I've made the mistake, here are the steps I've taken:

f(t) = (2-t)(t-15)

f(t) = t^2 - 17t + 30

f(t) = t^2 - 17t + (17/2)^2 + 30 - (17/2)^2

f(t) = (x - 17/2)^2 + 30 - (17/2)^2

f(t) = (x - 17/2)^2 + 120/4 - 289/4

f(t) = (x - 17/2)^2 - 169/4

The vertex form of the function tells me that the vertex should be located at (17/2, -169/4) however it turned out to be incorrect solution.

Now I'd assume that I can fully rely on the vertex form of the equation/function.

if I've made a mistake, PLEASE can someone tell me what I did incorrectly?

if I've not made a mistake, can someone explain PLEASE!(1 vote)- The error is when you changed
`(2-t)(t-15)`

into`t^2 - 17t + 30`

When you FOIL`(2-t)(t-15)`

, you get`2t - 30 - t^2 + 15t`

=`- t^2 + 17t - 30`

I think you treated (2-t) as (t-2) when you multiplied the binomials together, so all your signs are incorrect.

Hope this helps.(2 votes)

## Video transcript

- [Voiceover] So, I have three
different functions here. I know they're all called f, but we're gonna just assume
they are different functions. And, for each of these, I
want to do three things. I want to find the zeros. And so, the zeros are the input values that make the value of the
function equal to zero. So, here it'd be the t values
that make f of t equal zero. Here, it'd be the x values that make the function equal zero. So, I want to find the zeros. I also want to find the
coordinates of the vertex. And finally, I want to find the equation of the line of symmetry. Line of symmetry. And in particular, to make it
a little bit more specific, the vertical line of symmetry,
which will actually be the only line of symmetry for these three. So, pause the video and
see if you can figure out the zeros, the vertex,
and the line of symmetry. And I'm assuming you just did that and now I am going to attempt to do it. And if, at any point, you get inspired pause the video again and
keep on working on it. The best way to learn this
stuff is to do it yourself. So, let's see, so let's
first find the zeros. So, to find the zeros, we can set t minus five squared,
minus nine, equal to zero. So, we could say t ... for what t's does t minus five squared,
minus nine, equal zero? Let's see, to solve this, we
could add nine to both sides. And so we could say, if
we add nine to both sides the lefthand side's just
t minus five squared. The righthand side is going to be nine. And so, if t minus five squared
is nine, that means that t minus five could be equal to the positive square root of nine or t minus five could equal the negative square root of nine. And to solve for t, we could
add five to both sides, so we get t is equal to
eight or t is equal to, if we add five to both sides here, t is equal to two. And just like that, we
have found the zeros for this function because if
t is equal to eight or two, the function is going to be zero. F of eight is zero and f
of two is going to be zero. Now, let's find the vertex,
the coordinates of the vertex. The coordinates of the vertex. So, the x coordinate of the vertex, or sorry, I should say the
t coordinate of the vertex since the input variable here is t, the t coordinate of the
vertex is going to be halfway in between the zeros. It's going to be halfway in between where the parabola, in this case, is going to intersect the x axis, or the t axis, I keep saying x axis, the t axis for this case. So, halfway between eight and two. Well, it's gonna be the average of them. Eight plus two, over two. That's ten over two, that's five. The t coordinate is five and five is three away from eight
and three away from two. And, when t is equal to
five, what is f of t? What is f of five? Well, when t is equal to five, five minus five squared is just zero. And then, f of five is just
going to be negative nine. And this form of a function,
this is actually called vertex form because it's very
easy to pick out the vertex. It's very easy to
realize, like, okay look, for this particular one, we're
going to hit a minimum point when this part of the
expression is equal to zero because this thing, the
lowest value it can take on is zero cause you're squaring it, it can never take on a negative value. And it takes on zero
when t is equal to five. And when that's, if this part is zero, then the f of five is
going to be negative nine. So, just like that, we have
established the vertex. Now, we actually have a lot of information if we wanted to draw it. So, if we want to draw this function. I'll just do a very quick sketch of it. Whoops. So, a very quick sketch of it. So, that is our t axis, not our x axis, I have to keep reminding myself. And that is my, let's call that my y axis. And we're going to graph
y is equal to f of t. Well, we know the vertex is at the point five comma negative nine. So, this is, t is equal to five and y is equal to negative nine, so that's the vertex right over there. And then, we know we have zeros at t equals eight and t equals two. So, t equals eight and t equals two. Let me make that a little
bit in, t equals two. Those are the two zeros. So, eight and two. And so, just like that,
we can graph f of t, or we can graph y is equal to f of t. So, y is equal to f of t is
going to look something like let me draw it a little
... something like that That's the graph of y is equal to f of t. Now, the last thing that I
said is the line of symmetry. Well, the line of symmetry is going to be the vertical line that
goes through the vertex. So, the equation of that line of symmetry is going to be t is equal to five. That's really just the t
coordinate of the vertex. That defines the line of symmetry. Let's do the other two right over here. So, what are the zeros? Well, if you set this equal
to zero, if we say x plus two times x plus four is equal to zero, well that's going to happen
if x plus two is equal to zero or x plus four is equal to zero. This is going to happen if we
subtract two from both sides, when x is negative two,
and if we subtract four from both sides, or when x
is equal to negative four. As we said, the vertex, the
x coordinate of the vertex, is going to be halfway in between these. So, it's going to be negative
two plus negative four, over two, so that would
be negative six, over two, which is just negative three. Negative three. And when x is negative
three, f of x is going to be, let's see, it's going to be negative one times one, right? Negative three plus two is
negative one, and so times one. So, it's just going to be negative one. There you have it. And the line of symmetry is
going to be the vertical line x is equal to negative three. And once again, we can
graph that really fast. So, let me, this is my y axis. See everything is
happening for negative x's, so I'll draw it a little
bit more skewed this way. This is my x axis. And we see that we have zeros
at x equals negative two and x is equal to negative four. So, negative one, two, three, four. So, we have zeros there, negative two, be careful. Negative two and negative four. And the vertex is at negative
three comma negaitve one. So, negative three comma negative one. Make sure we see that, so this is negative one right over
here, negative one. This is negative two,
this is negative four. And so we can sketch out what the graph of y is equal to f of x
is going to look like. It's gonna look something like that. That is y is equal to f of x. Let's do one more. Hopefully, we're getting the hang of this. So, here, to solve x squared,
plus six x, plus eight is equal to zero, it will
be useful to factor this. And so, this can be written as, and if you have trouble doing this,
I encourage you to watch videos on factoring polynomials, What adds up to six and when you take their product is eight? Well, four and two. Four plus two is six and
four times two is eight. So, is equal to zero. And then, this is actually
the exact same thing as what we have in blue right over here. These are actually the
exact same function. They're just written in different forms. And so, the solutions are
gonna be the exact solutions that we just saw right
over here and the graph is going to be the same thing
that we have right over there. So, same vertex, same line
of symmetry, same zeros. These functions were just
written in different ways.