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## Get ready for Algebra 2

### Course: Get ready for Algebra 2>Unit 3

Lesson 11: Features and forms of quadratic functions

# 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
• If the coefficient of the squared term in the quadratic function is negative, then your parabola will face down and vice verse.
• 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.
• How do you figure out the x functions when trying to find the y values of a parabola?
• 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.
• 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
• 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.
• y=-(x-2)²+3
\$\$\$
• 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.
• For the question in the format of 'vertex form,' how do we get the zeros?
• 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.
• Are the zeros the x intercepts?
• Yes, provided the zeros are real numbers, then they are the x-intercepts for the parabola.
• How to write a functions that represents a graph
• 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?
• 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?
• 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`