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Please help and show work-example-1
User Xersiee
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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)

User Rinesse
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