Question

An expression is shown below:

f(x) = −16x2 + 24x + 16

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 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

#A

Take y =0

• y=-16x²+24x+16
• -16x²+24x+16=0
• -2x²+3x+2=0

On solving we will get

• x intercepts=(-0,5,0) and (2,0)

#B

Find x co ordinate of vertex

• x=-b/2a
• x=-24/-32
• x=3/4

Find y

• y=-16(3/4)²+24(3/4)+16
• y=25

As a is negative vertex is maximum as its facing downwards.

#C

Steps:-

• Put vertex on graph
• Put two x intercepts on graph .
• Draw a open hand parabola passing through three points

Answer 2

`\textsf{A)} \quad \left(-(1)/(2) , 0\right) \textsf{ and } (2, 0)`

`\textsf{B)} \quad \left((3)/(4),25 \right)`

C) see explanation

Explanation:

Given function:

`f(x) =-16x^2 + 24x + 16`

Part A

To find the x-intercepts, set the function to zero, factor and solve for x:

`\implies f(x)=0`

`\implies -16x^2 + 24x + 16=0`

`\implies -8(2x^2-3x-2)=0`

`\implies 2x^2-3x-2=0`

`\implies 2x^2-4x+x-2=0`

`\implies 2x(x-2)+1(x-2)=0`

`\implies (2x+1)(x-2)=0`

Therefore:

`\implies 2x+1=0 \implies x=-(1)/(2)`

`\implies x-2=0 \implies x=2`

Therefore, the x-intercepts of the graph of f(x) are

`\left(-(1)/(2) , 0\right) \textsf{ and } (2, 0)`

Part B

As the leading coefficient is negative, the parabola will open downwards. Therefore, the vertex of the graph of f(x) will be a maximum.

The x-coordinate of the vertex is the midpoint of the x-intercepts:

`\implies \sf midpoint=(2+\left(-(1)/(2)\right))/(2)=(3)/(4)`

To find the y-coordinate of the vertex, substitute the x-value into the function:

`\implies f\left((3)/(4)\right)=-16\left((3)/(4)\right)^2 + 24\left((3)/(4)\right) + 16`

`\implies f\left((3)/(4)\right)=-9+18+16`

`\implies f\left((3)/(4)\right)=25`

Therefore, the coordinates of the vertex are:

`\left((3)/(4),25 \right)`

Part C

Find the y-intercept by substituting x = 0 into the function:

`\implies f(0) =-16(0)^2 + 24(0) + 16=16`

Therefore, the y-intercept is (0, 16)

To graph f(x)

• Plot the vertex
• Plot the x-intercepts
• Plot the y-intercept
• Draw a parabola opening downwards with the vertex as the maximum point.

The axis of symmetry is the x-value of the vertex. Use this to help ensure the curve is symmetrical.