Quadratic Function

Zetrics

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For question 2. How do you know is maximum/minimum? And how to get 2x from the question?
 

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The TP form of a quadratic function is \(\displaystyle y=a(x-h)^2+k\) where (h, k) is the TP.

Your TP is (1, -2) so your equation is \(\displaystyle y=a(x-1)^2-2\).

You're also told that the graph goes through (0, -3), So put x=0 and y=-3 into your function and find a.

(If you draw a diagram labelling TP(1, -2) and the point (0, -3) you can see whether the parabola will have a max or min.)

There is no need to expand out the answer. Leave it in TP form as given in the answer you have.
 
I solved it differently than The Cat. The Cat's method is quicker than my method but not everyone thinks alike (usually I do not think like a cat, especially a smart cat) so I will show you my method. After seeing these methods if you come up with your own, then please post it.

Let \(\displaystyle y = ax^2 + bx + c\). Since (1,-2) is a turning point it must be on the quadratic along with the point (0,-3). This gives us 2 equations

\(\displaystyle -2 = a+b+c\) and \(\displaystyle -3 =c\). Combining yield that a+b = 1.

Now \(\displaystyle y' = 2ax + b\) and equals 0 when x=1 (the turning point). So 0 = 2a + b.

Solving a+b=1 and 2a+b =0 we get a=-1 and b= 2. Now we have a= -1, b= 2 and c=-3.

So the quadratic equation is \(\displaystyle y = -x^2 + 2x -3\)

Next you should complete the square and get \(\displaystyle y =-(x-1)^2 - 2\)

If you have any question please post back.
 
Jomo, your way assumes knowledge of calculus. You could overcome this by using x= -b/2a for TP.
 
I assume that when a post is placed in the advanced math section that the poster knows calculus …
That's a big assumption, when the post is about high-school math.

;)
 
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