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# pls help!!!

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165
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+10

Find area of the isosceles triangle formed by the vertex and the x-intercepts of parabola y=x^2-4x+21.

Jan 14, 2021

#1
+117576
+1

y = x^2   - 4x +  21

x coordinate of the  vertex      4 / (2)  =  2

y coordinate of the  vertex =   2^2  - 4(2)  + 21 =  17

The verxtex   is   ( 2,17).....above the x axis...since this parabola turns upward, there aren't any x intercepts...

However...if you mean   y   =  -x^2  -4x + 21

The  x coordinate of the  vertex is   4/ (2 * -1)  =   -2

And the y coordinate  of the vertex is  - (-2)^2  - 4(-2)  + 21  =  -4 + 8 + 21 =  25....this is the  height of the triangle

We can find the x intercepts as

-x^2 -4x +  21  =  0     multiply through by  -1

x^2  + 4x  - 21  =  0       tactor

(x   + 7)  ( x  - 3)  = 0

Set  each factor to 0 and solv for  x  and we  get  that x=   -7  and  x =  3

The  base of the  triangle  = ( 3 - - 7)   =10

So....the area is  (1/2) (base)(height)  = (1/2) (10)(25)  =   125

Jan 14, 2021
#2
0

thank you very much CPhill you the man

Guest Jan 14, 2021