Use the function f(x) to answer the questions f(x) = 5x^2 + 2x - 3
Part A: What are the x-intercepts of the graph of f(x)? Show your work (2 points)
Part B: Is the vertex of the graph of f(x) going to be a maximum or a minimum? What are the coordinates of the vertex? Justify your answers and show
your work (3 points)
Part 2: 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 (5 points)

9514 1404 393

  A: -1, 3/5

  B: (-1/5, -3 1/5)

  2: plot the given points and draw a curve through them.

Step-by-step explanation:

The x-intercepts are most easily found from a graph of the function. I like the Desmos graphing calculator for this purpose. The x-intercepts are -1 and 3/5.

You can also find the x-intercepts by factoring the function.

  f(x) = 5x^2 +5x -3x -3 = 5x(x +1) -3(x +1) = (5x -3)(x +1)

The x-intercepts are the values of x that make the factors zero:

  5x -3 = 0   ⇒   x = 3/5

  x +1 = 0   ⇒   x = -1

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The vertex is on the axis of symmetry, which is the vertical line at the value of x halfway between the zeros. (-1 +3/5)/2 = (-2/5)/2 = -1/5

The value of f(x) there is ...

  f(-1/5) = (5(-1/5) +2)(-1/5) -3 = (-1 +2)(-1/5) -3 = -1/5 -3 = -3 1/5

The vertex is (-1/5, -3 1/5).

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The easiest step to take to graph the function is to type it into a graphing calculator (see attached).

If you're graphing this by hand, the three points given in parts A and B will be on the graph. The leading coefficient in the function is 3, so the x^2 parent function is vertically stretched by a factor of 3. This can make it a little more difficult to find points that lie on the graph.

Recognizing that the vertex form equation is ...

  f(x) = 3(x +1/5) -3 1/5

I might choose x-values that are -1/5 ± 1/2 to get y-values that are -3 1/5 + 3/4. And, x = -1/5 ± 1 ⇒ y = -3 1/5 +3.

i think it's eleven too, don't woosh me here, but according to the "experts" it's 2.

the proof starts from the peano postulates, which define the natural

numbers n. n is the smallest set satisfying these postulates:

  p1.   1 is in n.

  p2.   if x is in n, then its "successor" x' is in n.

  p3.   there is no x such that x' = 1.

  p4.   if x isn't 1, then there is a y in n such that y' = x.

  p5.   if s is a subset of n, 1 is in s, and the implication

      (x in s => x' in s) holds, then s = n.

then you have to define addition recursively:

  def: let a and b be in n. if b = 1, then define a + b = a'

      (using p1 and p2). if b isn't 1, then let c' = b, with c in n

      (using p4), and define a + b = (a + c)'.

then you have to define 2:

  def:   2 = 1'

2 is in n by p1, p2, and the definition of 2.

theorem:   1 + 1 = 2

proof: use the first part of the definition of + with a = b = 1.

      then 1 + 1 = 1' = 2   q.e.d.

note: there is an alternate formulation of the peano postulates which

replaces 1 with 0 in p1, p3, p4, and p5. then you have to change the

definition of addition to this:

  def: let a and b be in n. if b = 0, then define a + b = a.

      if b isn't 0, then let c' = b, with c in n, and define

      a + b = (a + c)'.

you also have to define 1 = 0', and 2 = 1'. then the proof of the

theorem above is a little different:

proof: use the second part of the definition of + first:

      1 + 1 = (1 + 0)'

      now use the first part of the definition of + on the sum in

      parentheses:   1 + 1 = (1)' = 1' = 2   q.e.d.