Can I get some algebra help, polynomial functions? f(x) = -16x2 + 60x + 16

Question

anonymous · Answer

Part A: What are the x-intercepts of the graph of the f(x)? Show your work.

Part B: Is the vertex of the graph of f(x) going to be a maximum or minimum? What are the coordinates of the vertex? Justify your answers and show your work.

Part C: What are the steps you would use to graph f(x)? Justify that you can use the answers obtained in Part A and Part B to draw the graph.

anonymous · Answer

@ganeshie8 @saifoo.khan @SolomonZelman  @ash2326 @Abhisar

akashdeepdeb · Answer

Hi @gman957  Do you know what x-intercept of a function is? :)

anonymous · Answer

@AkashdeepDeb  yes

akashdeepdeb · Answer

Then, can you solve part A?

anonymous · Answer

No, I think I'm supposed to use Rational root thereom or something to solve this and I forgot how to solve normally. (these past lessons are all focused on factoring) @AkashdeepDeb

akashdeepdeb · Answer

The x intercept of anything is where the graph of the function cuts the x axis.
So where does any function $f(x)$ cut the x axis?
When the y value = 0 or when the function value = 0 the graph cuts the x axis.

So you have to factorize this function and then equate f(x) = 0 and find out for which 'x' the function'll be zero.

The 2 x's you find will be the x intercepts.

akashdeepdeb · Answer

My battery's going to run out, so here's the basic thing:

PART A : Basic factorization. Factorize so that you get something in this form: $$(x-a)(x-b) = 0$$
a,b will the be th x intercepts or the solutions of the function.

PART B : Since the function is a second degree polynomial, the function is a parabola. It looks a little something like this:

|dw:1406749986384:dw|

A few rules about a parabola: $$f(x) = ax^2 + bx + c$$
|dw:1406750041281:dw|

PART C : You can graph x by the following method.

Find out if the graph OPENS upward or downwards [ie. if a > 0 or a < 0]. You'd then get a basic idea of how the graph looks like. 

After that, find out the x intercepts and the y intercepts and mark them. And then draw a good-looking curve accordingly.

Hopefully, this helped. :)

anonymous · Answer

For part a I now have $$0 = -16x^2 + 6 4 - 4 + 16               $$

akashdeepdeb · Answer

Yes,
$$f(x) = -16x^2 + 60x + 16$$
$$f(x) = -16x^2 + 64x - 4x + 16$$
$$f(x) = -16x(x - 4) - 4(x-4)$$
$$f(x) = (-16x - 4)(x-4)$$
$$f(x) = -4(4x+1)(x-4)$$

Now find the zeroes, by equation the function to 0.

Because,

|dw:1406784422276:dw|

$$0 = -4(4x+1)(x-4)$$
$$x = \frac{-1}{4}, 4$$

So when x = 4, f(x) = 0 [Try putting the value of x in f(x) to check]
And also when x = $\frac{-1}{4}$, f(x) = 0 [Try this one too!]

So now, you got your x-intercepts, roots, solutions, zeroes or whatever.

Did you get this part? :)