Question

Which function has a vertex at (2, –9)? f(x) = –(x – 3)2 f(x) = (x + 8)2 f(x) = (x – 5)(x + 1) f(x) = –(x – 1)(x – 5

Answer 1

Answer:

the answer is C

Step-by-step explanation:

on edgenuty i just clicked C

Answer 2

You can use the vertex form of a quadratic equation to find out which of the given equation has got vertex at  (2, -9)

The function that has a vertex at (2,-9) is given by

Option C:  f(x) = (x – 5)(x + 1)

What is vertex form of a quadratic equation?

If a quadratic equation is written in the form

$y = a(x-h)^2 + k$

then it is called to be in vertex form. It is called so because when you plot this equation's graph, you will see vertex point(peak point) is on (h,k)

Using the above definition to find the function out of the given functions whose vertex would be at (2,-9):

First of all, i used this vertex formula thing since it is visible that there are quadratic functions. Otherwise, you had to use calculus to get critical points, then second derivative of functions to find the character of critical points as minima or maxima or saddle etc to get the location of vertex point.

Now,

Since we have (h,k) = (2,-9), thus, the vertex form of quadratic equation having this point as vertex would look like:

$f(x) = a(x-h)^2 + k = a(x-2)^2 + 9$

Since options have coefficient of $x^2$ as +1 or -1 (open the square thing and think of them in your mind. You will see x^2 will have no specified coefficient other than - sign (which is -1) which multiplies with it, and thus coefficient as 1 as 1 is always a factor), we will try both values for a

• Case 1: a = 1

Then we have:

$f(x) = a(x-2)^2 - 9 |_{a=1}\\\\f(x) = (x-2)^2 - 9 = x^2 - 4x +4 - 9 = x^2 - 4x - 5 = x^2 - 5x + x - 5 = x(x-5) +1(x-5) = (x-5)(x-1)\\\\f(x) = (x-5)(x+1)$

• Case 2: a = -1

Then we have:

$f(x) = a(x-2)^2 - 9 |_{a=-1}\\\\f(x) = -(x-2)^2 - 9 = -x^2 + 4x -4 - 9 = -x^2 + 4x - 13$

(not more easy factors visible as 13 is prime and we can't split 4 in integers to make monomial factor). It will come by experience. The more you do it, you will see it if its going to get expressed as easy factors or not. By easy factors, i mean factors which are simple looking, including only integers and no complex stuffs like roots or fractions etc.

Thus, only from case 1, we have the one of the options visible. That option is option C.

Thus,

The function that has a vertex at (2,-9) is given by

Option C:  f(x) = (x – 5)(x + 1)

Learn more about vertex form of quadratic equations here:

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