Proportion

A graph is symmetric about the $y$-axis if it comes from an even function. That means every input and its opposite give the same output.

Mathematically, we write: $f(x)=f(-x)$

So whatever happens at $x$ must match what happens at $-x$, creating a mirror image across the vertical axis.

Now apply the even-function condition to the power function $f(x)=k{x}^n$. To be symmetric about the $y$-axis, we need:

$k{x}^n=k(-x{)}^n$

Because $k>0$, we can divide both sides by $k$: $x^n=(-x{)}^n$

For this to be true for every real $x$, the exponent must make the sign disappear. That happens only when $n$ is an even integer (such as 2, 4, 6, ...).

To confirm, compare an even power (like $n=2$) with an odd power (like $n=3$) by evaluating at $x=2$ and $x=-2$. With an even power, the outputs match; with an odd power, they do not.

When $n$ is an odd integer, the sign stays negative: $(-x{)}^n=-(x^n)$ So $f(-x)=-f(x)$, which is symmetry about the origin (an odd function), not symmetry about the $y$-axis. So that choice should be removed.

If $n=0$, then $y=k{x}^0=k$, which is just a constant (a flat horizontal line), so it is not U-shaped.

If $n$ is negative (for example $-2$), then the function behaves like $y=k/x^2$, which has a vertical asymptote at $x=0$ and breaks into two branches rather than making one continuous U-shape through the origin.

With an even integer exponent, $y=k{x}^n$ gives the same (nonnegative) output for $x$ and $-x$. That creates the familiar U-shaped curve and perfect symmetry about the $y$-axis.

We are looking for the kind of exponent $n$ in $y=k{x}^n$ (with $k>0$) that makes the graph both U-shaped and symmetric about the $y$-axis. To make testing easier, take $k=1$, so we look at: $y=x^n$

Test the odd-integer option using values like $n=1$ and $n=3$.

• If $n=1$, then $y=x$, which is a straight line.
• If $n=3$, then $y=x^3$. For example, when $x=1$, $y=1$, but when $x=-1$, $y=-1$.

Since the outputs for $x$ and $-x$ are not equal, the graph is not mirrored in the $y$-axis. It has origin symmetry and an S-like shape, not a U.

Now substitute $n=0$:

$y=x^0=1\text{ (for }x\ne 0)$

This produces a horizontal line at $y=1$. It may be symmetric, but it is not U-shaped.

Test a negative exponent such as $n=-2$:

$y=x^{-2}=\frac{1}{x^2}$

As $x$ approaches $0$, the values grow without bound, and the graph splits into two separate parts on either side of the $y$-axis. That is not one continuous U-shaped curve.

Now test even integers such as $n=2$ and $n=4$.

• For $n=2$, $y=x^2$ gives the classic U-shaped parabola, and $(x{)}^2=(-x{)}^2$.
• For $n=4$, $y=x^4$ is also U-shaped and still has matching outputs for $x$ and $-x$.

These examples satisfy both conditions.

From the tested cases, only an even integer exponent consistently produces a U-shaped graph that is symmetric about the $y$-axis.

We are asked which $n$ makes the power graph $y=k{x}^n$ (with $k>0$):

1. Symmetric about the $y$-axis: the left and right sides match, so $f(x)=f(-x)$.
2. U-shaped: like a parabola or similar curves that open upward.

Think about common odd powers:

• $n=1$ gives $y=x$, a line through the origin.
• $n=3$ gives $y=x^3$, an S-shaped curve.

Odd-power graphs have rotational symmetry about the origin, not mirror symmetry across the $y$-axis, and they are not U-shaped. So that option can be discarded.

Consider the remaining exponent types:

• Zero ($n=0$): $y=k{x}^0=k$ is a constant line, not a U.
• Negative integers: for example $y=x^{-2}=\frac{1}{x^2}$. This is symmetric about the $y$-axis, but it has a vertical asymptote at $x=0$ and splits into two branches, so it is not a single continuous U-shape through the origin.

So both of these can be eliminated.

Now think of even powers such as $n=2$ and $n=4$:

• With $n=2$, $y=k{x}^2$ is a parabola: U-shaped and symmetric because $(-x{)}^2=x^2$.
• With $n=4$ (and other even integers), the curve stays U-shaped and keeps the same left-right symmetry, though the bottom shape changes slightly as $n$ increases.

This fits the description exactly.

Graph a few examples on a GDC (for instance, $n=2$, $n=3$, and $n=-2$) to see how changing $n$ changes symmetry and whether the curve is a single U-shape.

From recognizing standard power-function shapes, the exponents that produce a U-shaped curve symmetric about the $y$-axis are even integers.

Because the problem says $y$ is directly proportional to $x$, we model the relationship using a direct proportion equation.

In Sequences, Formulae and Proportional Reasoning, direct proportion is expressed as:

$y=kx$

The number $k$ is the constant of proportionality, meaning it stays fixed for every matching pair of $x$ and $y$ values.

Use the values provided ($y=32$ when $x=4$) and substitute into $y=kx$:

$\begin{array}{rl}32& =k\times 4\\ k& =\frac{32}{4}\\ k& =8\end{array}$

Now the rule is $y=8x$. Substitute $y=56$ to solve for $x$:

$\begin{array}{rl}56& =8\times x\\ x& =\frac{56}{8}\\ x& =7\end{array}$

Compare with the answer choices:

• Option D ($448$) comes from doing $56\times 8$, but to isolate $x$ in $56=8x$ you must divide by $8$.
• Option C ($8$) is the constant $k$, not the required $x$ value.
• Option B ($14$) could happen from an incorrect division or mixing up the proportional relationship.

So the correct value is $x=7$.

If $y$ is directly proportional to $x$, they increase (or decrease) together at a constant rate. That means the ratio stays the same for any two matching pairs:

$\frac{y_1}{x_1}=\frac{y_2}{x_2}$

This property from Sequences, Formulae and Proportional Reasoning lets you solve quickly without separately writing $k$ first.

From the information given:

• $y_1=32$
• $x_1=4$
• $y_2=56$
• $x_2=x$

Put these into the ratio equality:

$\frac{32}{4}=\frac{56}{x}$

First simplify the left-hand side:

$\frac{32}{4}=8$

So the equation becomes:

$8=\frac{56}{x}$

Now solve for $x$ by multiplying by $x$ and then dividing by $8$:

$\begin{array}{rl}8x& =56\\ x& =\frac{56}{8}\\ x& =7\end{array}$

Check the distractors:

• Option D: This is $56\times 8$, which can happen if you multiply instead of keeping the ratios equal.
• Option C: $8$ represents the constant ratio (or $k$), not the unknown $x$.
• Option B: $14$ is twice the correct value, likely from an arithmetic slip when dividing.

Using the ratio equality, we get $x=7$ when $y=56$, which matches Option A.

Direct proportion means $y$ changes at a constant rate compared with $x$. So the quotient of the two values is fixed, and we call it $k$:

$\frac{y}{x}=k$

Use the given pair $y=32$ when $x=4$ to calculate $k$:

$\begin{array}{rl}k& =\frac{32}{4}\\ k& =8\end{array}$

So any correct pair $(x,y)$ in this question must satisfy $\frac{y}{x}=8$.

Now check each option by seeing whether $\frac{56}{x}$ equals $8$.

Try Option D ($x=448$): $\frac{56}{448}=0.125$

Because $0.125\ne 8$, Option D cannot be correct.

Try Option B ($x=14$): $\frac{56}{14}=4$

Since $4\ne 8$, this choice is wrong.

Try Option C ($x=8$): $\frac{56}{8}=7$

Because $7\ne 8$, Option C does not work.

Try Option A ($x=7$): $\frac{56}{7}=8$

This matches the constant ratio, so $x=7$ is the correct answer.

We need the geometric meaning of the constant $k$ in the direct proportion rule:

$y=kx$

This rule graphs as a straight line that goes through the origin. To interpret $k$, we can match this equation to a familiar straight-line form from Coordinate Geometry.

A common way to write a straight-line equation is:

$y=mx+c$

In this form:

• $m$ is the gradient (slope/steepness).
• $c$ is the $y$-intercept (the point where the line meets the $y$-axis).

Place the two equations side-by-side:

$\begin{array}{rl}y& =kx\\ y& =mx+c\end{array}$

To compare directly, rewrite the direct proportion as:

$y=kx+0$

Now it’s clear that:

• $k$ plays the same role as $m$ (the gradient).
• $c=0$, meaning the line’s $y$-intercept is at the origin.

If you graph several lines of the form $y=kx$ with different values of $k$, the line always goes through $(0,0)$, but it becomes more or less steep. That change in steepness is exactly what the gradient describes.

From the comparison:

• The $x$-intercept and $y$-intercept are both $0$ because the line goes through the origin, but that is not the job of $k$.
• Curvature describes bending, but this graph is a straight line, so curvature is not relevant.

Because $k$ is in the same position as $m$ in $y=mx+c$, it represents the line’s steepness, i.e., the gradient.

We are given $y=kx$, which is a linear direct proportion. To decide what $k$ means geometrically, we can eliminate options by using what we know about this type of graph.

This connects to Coordinate Geometry and Linear Modelling, where recognizing line features from equations is essential.

Find the intercept at $x=0$:

$\begin{array}{rl}y& =k(0)\\ y& =0\end{array}$

So the graph always passes through $(0,0)$. The $x$-intercept and $y$-intercept are fixed at $0$ no matter what $k$ is, so $k$ cannot represent either intercept.

The equation $y=kx$ is first-degree (the highest power of $x$ is 1), which produces a straight line.

A straight line does not curve, so $k$ cannot be describing curvature.

After removing intercepts and curvature, the remaining idea is the gradient. We can also verify by comparing with:

$y=mx+c$

Here $c=0$ and $k$ is in the same position as $m$. Since $m$ is the gradient, $k$ must be the gradient.

With the other options ruled out, $k$ must represent the line’s steepness (its gradient).

The relationship is a direct proportion relationship, written as $y=kx$. Its graph is a straight line and it always goes through $(0,0)$, which is a key idea in Coordinate Geometry and Linear Modelling.

To see what $k$ stands for, rearrange by dividing by $x$:

$k=\frac{y}{x}$

So $k$ is a constant ratio comparing $y$ to $x$ for points on the line.

The gradient is defined as rise over run, meaning change in $y$ divided by change in $x$:

$\text{gradient}=\frac{\Delta y}{\Delta x}$

Because the line goes through $(0,0)$, the gradient from the origin to a point $(x,y)$ is:

$\text{gradient}=\frac{y-0}{x-0}=\frac{y}{x}$

Since we also have $k=\frac{y}{x}$, the constant $k$ matches the gradient.

Consider the remaining options:

• The x-intercept and y-intercept are axis-crossing points. For $y=kx$, both are always at $0$, regardless of $k$.
• Curvature is about bending, but a linear graph is straight, so it has no curvature.

Because $k$ equals the rise/run ratio, it is exactly the gradient of the line.