Question

50 POINTS

Use the function f(x) to answer the questions.

f(x) = −16x2 + 22x + 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 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. (5 points)

Answer 1

work and answers below

Answer 2

`\text{Part A.}\\(-(1)/(8),0),\\((3)/(2),0)\\\\\text{Part B.}\\((11)/(16),(169)/(16))\\\\\text{Part C.}`

Draw a parabola concave down with vertex at
`((11)/(16),(169)/(16))`. Since the leading coefficient of the equation is -16, the parabola should appear thinner than its parent function
`y=x^2`. Ensure that the parabola passes through the points
`(\(-(1)/(8),0)` and
`((3)/(2),0)`.

Explanation:

Part A:

The x-intercepts of a function occur at
`y=0`. Therefore, let
`y=0` and solve for all values of
`x`:

`0=-16x^2+22x+3`

The quadratic formula states that the real and nonreal solutions to a quadratic in standard form
`ax^2+bx+c` is equal to
`x=(-b\pm √(b^2-4ac))/(2a)`.

In
`-16x^2+22x+3`, assign:

• 
  `a\implies -16`
• 
  `b\implies 22`
• 
  `c\implies 3`

Therefore, the solutions to this quadratic are:

`x=(-22\pm√(22^2-4(-16)(3)))/(2(-16)),\\x=(-22\pm 26)/(-32),\\\begin{cases}x=(-22+26)/(-32)=(4)/(-32)=\boxed{-(1)/(8)},\\x=(-22-26)/(-32)=(-48)/(-32)=\boxed{(3)/(2)}\end{cases}`

The x-intercepts are then
`\boxed{(-(1)/(8),0)}` and
`\boxed{((3)/(2),0)}`.

Part B:

The a-term is negative and therefore the parabola is concave down. Thus, the vertex will be the maximum of the graph. The x-coordinate of the vertex of a quadratic in standard form
`ax^2+bx+c` is equal to
`x=(-b)/(2a)`. Using the same variables we assigned earlier, we get:

`x=(-22)/(2(-16))=(-22)/(-32)=(11)/(16)`

Substitute this into the equation of the parabola to get the y-value:

`f(11/16)=-16(11/16)^2+22(11/16)+3,\\f(11/16)=(169)/(16)`

Therefore, the vertex of the parabola is located at
`\boxed{((11)/(16),(169)/(16))}`