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vertex of the parabola #1

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Hi, Carlos.
 
There are two ways to approach this problem.  We need to put this equation into a different form.
 
f(x) = a(x-h)2 + k, so the vertex will be the point (h, k).
 
One way is to complete the square:  f(x) = (x2 - 8x +    ) + 23.
 
What will make the perfect square?  Take half of 8, square it , and add it to the inside and subtract from the outside on the parenthesis.
 
This results in f(x) = (x2 - 8x + 16) + 23 - 16.
 
And f(x) = (x - 4)2 + 7.   Therefore, the axis of symmetry is x = 4, and the vertex is the point (4, 7)
 
Alternatively, you can find the value of f, by using the equation x = -b/(2a), where b is the coefficient of x and a is the coefficient on x2.  This results in x = -(-8)/(2*1).
 
So the x value of the vertex is 4 and the y value can be found by finding f(4).
 
f(4) = 42 - 8*4 + 23 = 16 - 32 + 23 = 7.  So the vertex is (4, 7).
 
I hope this helps.