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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.

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• 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.
• how did he find the y-vertex?in 2 equation
• 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?