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The graph of y=f(x) has the line x=5 as an axis of symmetry. The graph also passes through the point (-8, -7) Find another point that must lie on the graph of y=f(x).

 Jul 23, 2021
 #1
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Assuming a monic quadratic polynomial ....we  must  have  the point  (5, k)  as  the  vertex

 

Since  (-8, -7)  is on the  graph, we  have  that

 

-7  = ( -8 - 5)^2 +  k

 

-7  = (-13)^2  + k

 

-7  = 169 + k

 

k =  -7 -169  =  -176

 

So   the  function is

 

y=  (x - 5)^2  - 176

 

So......the point (5, -176)  also must be on the graph

 

 

cool cool cool

 Jul 23, 2021
edited by CPhill  Jul 23, 2021

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