Please help with #29, show work to understand, Thank you For Rach quacdvatic funchion, idlentify the vertex, axis ofSymmetny anda X- andy- intercepts. Then, aqraph thefunction. Determine the' domain andrange.+21F(x) = (x+1)2 -4y= (x-4)² - 22529-(x+1)? -5%3Dgx) - x² -1|#33)O gla) = 3(x +2)*+5+37%3D

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Description: Please help with #29, show work to understand, Thank you For Rach quacdvatic funchion, idlentify the vertex, axis ofSymmetny anda X- andy- intercepts. Then, aqraph thefunction. Determine the' domain andrange.+21F(x) = (x+1)2 -4y= (x-4)² - 22529-(x+1)

Answered: 29 y= = -(x+1)²-5

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter3: Functions And Graphs

Section: Chapter Questions

Problem 14T

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Please help with #29, show work to understand, Thank you

For Rach quacdvatic funchion, idlentify the vertex, axis of
Symmetny anda X- andy- intercepts. Then, aqraph the
function. Determine the' domain and
range.
+21
F(x) = (x+1)2 -4
y= (x-4)² - 2
25
29
-(x+1)? -5
%3D
gx) - x² -1|
#33)
O gla) = 3(x +2)*+5
+37
%3D

Expert Solution

Step 1

Given equation,

$y = - {(x + 1)}^{2} - 5$ ...(1)

Step 2

The given equation compare with the equation $y = a {(x - h)}^{2} + k$

Where $(h, k) =$ vertex

And x=h is the axis of symmetry

$a = - 1 h = - 1 k = - 5$

Thus the vertex of the given equation is $(- 1, - 5)$

And axis of symmetry is x = -5

Step 3

For the x-intercept put y = 0 in equation (1)

$0 = - {(x + 1)}^{2} - 5 \Rightarrow {(x + 1)}^{2} + 5 = 0 \Rightarrow x^{2} + 1 + 2 x + 5 = 0 \Rightarrow x^{2} + 2 x + 6 = 0 \Rightarrow x = \frac{- 2 \pm \sqrt{2^{2} - 4 (6)}}{2 (1)} \Rightarrow x = \frac{- 2 \pm \sqrt{4 - 24}}{2} \Rightarrow x = \frac{- 2 \pm \sqrt{- 20}}{2} \Rightarrow x = \frac{- 2 \pm 2 \sqrt{- 5}}{2} \Rightarrow x = - 1 \pm 5 i i s c o m p l e x n u m b e r .$

So, it has no x-intercept.

And for the y-intercept put x= 0 in (1)

$y = - {(0 + 1)}^{2} - 5 y = - 1 - 5 y = - 6$

So, y-intercept: $(0, - 6)$

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