# Thread: How to find f(x) by just a graph?

1. ## How to find f(x) by just a graph?

My question is on here : SparkNotes: SAT Subject Test: Math Level 2: Review Questions

#5. I don't get why and I don't get its explanations

Sorry for just pasting the website, I cannot copy the diagram. Thanks in advance!!

2. Hello, fabxx!

I don't get its explanations either . . .

5. A portion of the graph of $y = f(x)$ is shown.
Code:
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*           | *
*   *              *
*          |    *
*        *|                 *
- - - -*- - -*-+- - -*- - - - - * - - -
*    |       *      *
|           *
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Which of the following could be $f(x)$ ?

Note that the graph has four x-intercepts, two positive and two negative.

$(A)\;x^4 + ax^3 + bx^2 + cx$ .No
One of the x-intercepts is 0; this graph would contain the origin.

$(C)\;-x^4+ax^3 + bx^2 + cx + d$ .No
This quartic opens down.

$(D)\;x^3 + ax^2 + bx + c$ .No
This is a cubic function.

$(E)\;x^4+b^2x^2+c^2$ .No
To find the x-intercepts: . $x^4 + b^2x + c^2 \:=\:0$

Quadratic Formula: . $x^2 \;=\;\frac{-b^2 \pm\sqrt{b^4-4c^2}}{2}$

It can be shown that the right side is either negative or zero.

. . If it is negative, there are no real values of $x.$

. . If it is zero, there is only one x-intercept.

$(B)\;x^4+ax^2+b$ .Yes

To find the x-intercepts: . $x^4 + ax^2 + b \:=\:0$

Quadratic Formula: . $x^2 \;=\;\frac{-a \pm\sqrt{a^2-4b}}{2}$

If $a$ is negative, the right side is positive.

Hence, for $a < 0$, there are four x-intercepts:

. . $x \;=\;\pm\sqrt{\frac{-a \pm\sqrt{a^2-4n}}{2}}$

3. But the function in B is an even function while the graph is not that of an even function. So as written there is no answer.

Look at question #6. The given graph is not an odd function either.

That is just a poor set of problems and they are selling them as SAT review.

4. HAHAH thanks for reply Soroban. And also Plato!! haha