MHB Even Functions, Symmetry, Inverse Functions

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The discussion centers on identifying characteristics of an even function based on its graph, specifically focusing on coefficients a, b, c, d, and e. It is established that for the function to be even and symmetric about the y-axis, coefficients b and d must be zero. The analysis concludes that a must be negative, e must be negative due to the y-intercept being below zero, and c must be positive to ensure the function is above zero in certain intervals. Additionally, it clarifies that a function cannot be symmetric about the x-axis, as this would violate the definition of a function having a single output for each input. The conversation highlights the importance of understanding symmetry and the implications for the coefficients in polynomial functions.
confusedatmath
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Can someone explain why the answer is D

a < 0 because it finishes downwards
e < O because the y-intercept is in the negatives.
b, & d = zero (but i don't get this)
c is supposedly > 0 (nor do i get this)

According to the solutions the graph is an even function, and symmetrical about the y-axis/x-axis. I haven't studied this, can someone please explain for values b,d,c.
 

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Looking at the graph, we can see it is reflected across the $y$-axis, meaning it is an even function, which means:

$$y(-x)=y(x)$$

This means the coefficients of the terms having an odd exponent on $x$ must be zero, i.e.:

$$b=d=0$$

This leaves us with:

$$y(x)=ax^4+cx^2+e$$

Now, we see that:

$$y(0)=e<0$$

We also see that as $x\to\pm\infty$ we have $$y(x)\to-\infty$$. This means the dominant term, that is, $ax^4\to-\infty$ which implies $a<0$. But in order to have $y(x)>0$ on the two shown intervals, we must have $0<c$.
 
is there an equation for symmetry in x-axis?
 
confusedatmath said:
is there an equation for symmetry in x-axis?

A function, by definition, cannot be symmetric across the $x$-axis. For any given input, a function can only have one output.
 
Hello, confusedatmath!

Can someone explain why the answer is D

a < 0 because it finishes downwards
e < O because the y-intercept is in the negatives.
b, & d = zero (but i don't get this)
c is supposedly > 0 (nor do i get this)
We have: .y \:=\:ax^4 + bx^3 + cx^2 + dx + e

The graph is symmetric to the y-axis.

On the right, we have two x-intercepts, p and q.
On the left, we have two x-intercepts, \text{-}p and \text{-}q.

The function has the form:
. . (x-p)(x+p)(x-q)(x+q) \:=\:x^4 - (p^2+q^2)x^2 + p^2q^2

Since a is negative, we have:
. . y \:=\:-\left[x^4-(p^2+q^2)x^2 + p^2q^2\right]
. . y \:=\:-x^4 + (p^2+q^2)x^2 - p^2q^2\begin{array}{cc}\text{Therefore:} &amp; a\text{ neg.} \\ &amp; b = 0 \\ &amp; c\text{ pos.} \\ &amp; d = 0 \\ &amp; e\text{ neg.} \end{array}
 
wow mindblown. :O that was such a cool way of solving it! THANK YOU (Inlove)
 
confusedatmath said:
is there an equation for symmetry in x-axis?
A non-function relation such as x^2+ y^2= 1 of x^2+ (y- 2)^2= 1, may be "symmetric about the x-axis". Just as the test for "symmetry about the y-axis" is f(x)= f(-x) (so that replacing x with -x does not change the equation) so the test for "symmetry about the x-axis" is that replacing y with -y does not change the equation.

Because (-a)^2= a^2, in the first example above, x^2+ (-y)^2= x^2+ y^2= 1, the same as the original equation, so that is "symmetric about the x-axis". (Its graph is a circle about the origin.)

The second would give x^2+ (-y- 2)^2= x^2+ (-(y+ 1))^2= x^2+ (y+1)^2= 1 which is NOT the same as the original. The graph of x^2+ (y- 2)^2= 1 is a circle with center at (0, 2) so is symmetric about y= 2, NOT y= 0 which is the x-axis.
 

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