Explanation:
1) y=1
Obviously this is a line with a constant ordinate and parallel to the x-axis, it admits no x-intercepts since it cannot intersect the x-axis
The y-intercept is the point of coordinates (0,1)
Find the derivative of the function and solve it equal to zero to find the coordinates of the vertex
y=1
y'=1 =>0=1 Impossible
So it admits no vertex.
2)
x=3
Similarly this is a line with a constant abscissa and parallel to the y-axis.It does not have a y-intercept since it cannot intersect the y-axis.
The x intercept is the point of coordinates (1,0)
This function admits no derivative since it is a function independent of the x value (constant x)
Therefore it has no intercepts.
3)
f(x)=3x+2
Solve for f(0) to find the y-intercept
f(0)=3(0)+2=2
X-intercept (2,0)
Solve f(x)=0 to find the x-intercept(s)
3x+2=0
3x=-2
x=-2/3
Y-intercept (-2/3,0)
Find the derivative and solve it equal zero to find the coordinates of the vertex:
f(x)=3x+2
f'(x)=3
3=0 impossible, so this function admits no vertex.
4)
f(x)=x²+3x+4
to find the y-intercept solve x=0 or in other words, find f(0)
f(0)=0²+3(0)+4=4
y-intercept is (0,4)
to find the x-intercept(s) solve f(x)=0
x²+3x+4=0
find the roots to this equation
∆=b²-4ac
∆=3²-4(1)(4)
∆=9-16
∆=-7<0
which implies this function admits no x-intercepts
5)
f(x)=-x²+4x+16
f(0)=-(0)²+4(0)+16
f(0)=16
Y-intercept (0,16)
f(x)=0
-x²+4x+16=0
""Solve using the quadratic formula'"'
You get the roots:
x¹=-2.47 x²=6.47
x-intercepts (-2.47,0) (6.47,0)
f'(x)=-2x+4
f'(x)=0
-2x+4=0
-2x=-4
x=2
Find f(2)=-(2)²+4(2)+16=-4+8+16=20
Vertex (2,20)
6)
f(x)=-2x²+x+2
f(0)=-2(0)²+0+2=2
Y-intercept (0,2)
f(x)=0
-2x²+x+2=0 'use the quadratic formula'
x¹=-0.78 x²=1.78
x-intercepts (-0.78,0) (1.78,0)
f'(x)=-4x+1
f'(x)=0
-4x+1=0
-4x=-1
x=1/4=0.25
f(0.25)=-2(0.25)²+0.25+2=2.125
Vertex (0.25,2.125)