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In the following exercise a) write the equation in standard form b) use properties of standard form to graph the parabola:

y=-x2+12x-35

Short Answer

Expert verified

The equation in standard form can be given asy=-(x-6)2+1

The graph of the parabola can be given as

Step by step solution

01

Given information

We are given a equationy=-x2+12x-35

02

Write the equation in standard form

We have,

y=-x2+12x-35y=-1(x2-12x)-35y=-1(x2-12x+36)+36-35y=-(x-6)2+1

03

Write the axis of symmetry and find the vertex of parabola

The axis of symmetry can be given as x=h

we get

The vertex of parabola isrole="math" localid="1645503068309" (h,k)=(6,1)

04

Find y-intercept

Put x=0

role="math" localid="1645503109752" y=-(0-6)2+1y=36+1y=37

The coordinate is(0,37)

05

Find the point of symmetry (0,37)across the axis of symmetry

The point can be given as(12,-35)

06

find the x-intercept

put y=0

we get,

-(x-6)2+1=0(x-6)2=1 x-6=1x=5orx=7

07

Plot the graph of the parabola

The graph can be given as

08

Conclusion

The equation of the parabola can be given as y=-(x-6)2+1

The graph can be given as

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