Given function: [tex]f(x) = -16x^2 + 24x + 16[/tex]
Part A
To find the x-intercepts, set f(x) = 0
[tex]\sf -16x^2 + 24x + 16 = 0[/tex]
[tex]\sf -2x^2 + 3x +2 = 0[/tex]
[tex]\sf -2x^2 + 4x-x +2 = 0[/tex]
[tex]\sf -2x(x -2)-1(x-2) = 0[/tex]
[tex]\sf (-2x-1)(x-2) = 0[/tex]
[tex]\sf -2x-1=0, \ x-2 = 0[/tex]
[tex]\sf -2x=1, \ x=2[/tex]
[tex]\bold{ x=-\dfrac{1}{2}, \ x=2}[/tex]
Part B
To find the vertex, differentiate the function and set it (dy/dx) to 0
[tex]\sf \dfrac{dy}{dx}= \dfrac{d}{dx}(-16x^2 + 24x+16) = -32x+24[/tex]
[tex]\sf -32x + 24 = 0[/tex]
[tex]\sf x = -24/-32 = 3/4[/tex]
[tex]\sf y= -16(\dfrac{3}{4})^2 + 24(\dfrac{3}{4}) + 16 = 25[/tex]
Vertex: (0.75, 25)
To find if this is maximum or minimum. For a quadratic function like
ax^2 + bx + c. If a < 0 then maximum and if a > 0 then minimum.
Here out a = -16 which is < 0 so maximum.
Part C
Find y intercept by setting x = 0 into equation
y = -16(0)^2 + 24(0) + 16 = 16. So y-intercept: (0, 16)
Plot points of the x intercepts and the vertex on the graph and make a smooth curved graph. Your graph should look something like this:
