y = (x - 3 )^2 +4 please help solve by identify the minimum or maximum y value
That's a parabola. If you multiply it out, you'll see that the coefficient on the \(x^2\) term is positive, which means the parabola opens up (looks like a bowl). That means the vertex is the minimum point. You conveniently have the formula in what is called vertex form: \[y =a (x-h)^2+k\] where the vertex is at \((h,k)\). \(a = 1\) in this case. Where is the vertex?
If you had the formula in expanded or standard form: \[y = ax^2+bx+c\]you can find the x coordinate of the vertex by \[x = -\frac{b}{2a}\]and plug that value of \(x\) into the formula to find the y coordinate of the vertex. Don't need to know that to do this problem, but it beats factoring to get vertex form if you don't already have it.
I am not sure
The x^2 value is always going to be positive, as the parabola opens upwards and thus will have a minimum value, so it would need to equal 0 for the minimum y value.
Compare your equation with mine: \[y = (x-3)^2+4\]\[y=(x-h)^2+k\] what are the values of \(h\) and \(k\)?
please give me a moment to write out
sure, I'm not going anywhere...
I don't know I am having a hard time figuring out algebra
I have a tutor that sits with me once a week an I just can seem to grab on to algebra
are you there?
yes, just got back.
\[y=(x-h)^2+k\]\[y=(x-3)^2+4\] Look at the corresponding spots in each equation. Let's find the value of \(k\) first. There are 3 "chunks" in the second equation: \[y\]\[(x-3)^2\]\[4\] which of those most resembles \(k\) in the first equation? \(4\), right?
yes
Okay, now let's find \(h\). Any ideas?
h would be 3 right?
yes! (sorry about the delay, my family is bugging me for help on other projects)
understand
So our vertex is at \((h,k)\) and as we found that \(h = 3, k = 4\), that means the vertex is at \((3,4)\)
There's a nice page talking about parabolas and their formulas here: http://www.mathwarehouse.com/geometry/parabola/standard-and-vertex-form.php
okay thank you
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