The vertex will be when the velocity is equal to zero:
df/dx=-12x+24 (using the power rule for differentiation)
df/dx=0 only when:
-12x+24=0 add 12x to both sides
12x=24 divide both sides by 12
x=2, we find the y value to be:
y(2)=-6(2^2)+24*2-20
y(2)=-24+48-20=4
So the vertex is the point (2,4)
And the axis of symmetry is the line x=2
Now if you do not yet do calculus...
The vertex will occur midway between the two zeros.
6x^2-24x+20=0 (using the Quadratic Formula for simplicity)
x=(24±√96)/12
x≈(1.1835, 2.8165)
Now the vertex occurs at the average of the zeros...
x=(1.1835+2.8165)/2=2 (as we saw earlier)
y(2)=4
Vertex is at (2,4) and axis of symmetry is the vertical line x=2
<em>The point where a function (for our problem its the cubic function shown)</em>
<u>and/or</u>
<em>is a zero of the function.</em>
From the graph shown, we can clearly see that it cuts the x-axis at -1 and touches the x-axis at 2.
So the zeros are at -1 and 2.
ANSWER: {-1,2}
Standard form (which I assume is what they want you to turn these into) is y=mx+b.
12x-5y=20 is y=2.4x-4.
Subtract 12x from both sides, -5y=-12x+20.
Multiply both sides by -1, 5y=12x-20.
Divide both sides by 5, y=2.4x-4.
y=x+4 is already in standard form- I'm not quite sure what to do.
3x+y=3 is y=-3x+3.
Subtract 3x from both sides, y=-3x+3.
x=-y+3 is y=-x+3
Ad y to both sides, y+x=3.
Subtract x from both sides, y=-x+3.
Answer:
124°+95°+124°+54°+R = 180°
R = 143°
