Transforming Polynomial Graphs

Let $P(x) = x^2$, so the graph has its lowest point at $x = 0$. Now consider $y = P(x - 5) = (x-5)^2$. What value of $x$ makes this new expression equal to zero, and therefore where is the new lowest point? Say what the bracket had to be equal to, and what $x$ had to be for that to happen.

Four transformations, and the crucial split is outside the function against inside it. Changes written outside, such as $+c$ or a multiplier, do what they look like they do. A change written inside the brackets acts on the input, and therefore appears reversed.

Adding a constant to the output raises every point by the same amount.

$$y = P(x) + c \quad \text{translates the graph } c \text{ units up}$$

A negative $c$ moves it down. This one behaves exactly as it reads, because the addition happens after $P$ has finished: whatever height the original curve reached at a given $x$, the new curve reaches $c$ more.

Effects on the key features:

the $y$-intercept becomes $P(0) + c$;
the zeroes usually change completely, because the curve has been lifted off the axis;
the shape and the end behaviour are unchanged.

Take $P(x) = x^2 - 4$, with zeroes at $\pm 2$. Then $P(x) + 4 = x^2$ has a single zero at $0$, and $P(x) + 5 = x^2 + 1$ has none at all. A vertical shift can destroy zeroes or create them, which is exactly what raising or lowering a curve past the axis should do.

This is the one that looks backwards.

$$y = P(x-a) \quad \text{translates the graph } a \text{ units to the RIGHT}$$

So $y = P(x-5)$ moves right by $5$, and $y = P(x+3)$ moves left by $3$.

Why. The change is inside the brackets, so it acts on the input before $P$ ever sees it. For the new function to do what $P$ did at some value, the bracket must equal that value. If $P$ had a zero at $x = 2$, the new function is zero when $x - 5 = 2$, that is at $x = 7$. The input has to be made larger by $5$ to compensate for the subtraction, and larger means further right.

Effects on the key features:

every zero moves right by $a$;
the $y$-intercept becomes $P(-a)$, which usually changes;
the shape and end behaviour are unchanged.

Multiplying the output by $-1$ sends every point to the opposite side of the horizontal axis, at the same distance.

$$y = -P(x) \quad \text{reflects the graph in the horizontal axis}$$

Effects on the key features:

every zero stays exactly where it was, since $-0 = 0$;
the $y$-intercept becomes $-P(0)$;
both ends flip: a curve that rose to the right now falls;
maxima become minima and the reverse.

The zeroes being invariant is worth pausing on. A point on the axis has height zero, and multiplying zero by $-1$ leaves it at zero, so it does not move. The points where the curve meets the axis are precisely the points a vertical reflection cannot shift.

In factored form the reflection is visible directly: $y = (x-1)(x-4)$ becomes $y = -(x-1)(x-4)$, and the brackets, which carry the zeroes, are untouched.

Multiplying the output by a constant $k$ stretches the graph away from the horizontal axis when $|k| > 1$, and squashes it towards the axis when $0 < |k| < 1$.

$$y = kP(x) \quad \text{stretches vertically by a factor of } k$$

Every height is multiplied by $k$, so a point at height $3$ moves to height $3k$. A negative $k$ combines the stretch with a reflection, since $kP(x) = |k| \times (-P(x))$ when $k < 0$.

Effects on the key features:

every zero stays where it was, for the same reason as a reflection;
the $y$-intercept becomes $kP(0)$;
the ends keep their directions if $k > 0$ and swap if $k < 0$;
the leading coefficient is multiplied by $k$.

This is why Lessons 9 and 10 kept saying that zeroes alone do not determine a polynomial. Every $y = k(x-1)(x-4)$ shares the same two crossings; they differ by exactly this vertical stretch, and one extra point is needed to fix $k$.

Written together the four appear as

$$y = k\,P(x - a) + c$$

and applying them in the right order matters. Working from the inside out:

1. $x - a$: translate horizontally by $a$.
2. multiply by $k$: stretch vertically, reflecting as well if $k < 0$.
3. $+c$: translate vertically by $c$.

The order of steps 2 and 3 is not interchangeable. Stretching by $2$ and then raising by $3$ sends a point at height $1$ to $2 \times 1 + 3 = 5$; raising first and then stretching sends it to $2(1+3) = 8$. The equation $y = kP(x-a) + c$ says stretch first, because in the expression the multiplication is applied before the addition.

A single check catches almost every slip. Take one point you know on $y = P(x)$ and push it through the steps by hand, then confirm your transformed sketch passes through the result.

Five items on the four transformations. Work each one, then reveal the answer.

1. 1 Describe the change from $y = P(x)$ to $y = P(x) - 7$.

   Show answer
   Outside the function, so it behaves as it reads.Translate $7$ units down

2. 2 Describe the change from $y = P(x)$ to $y = P(x+4)$.

   Show answer
   Inside the brackets, so the effect is reversed.Translate $4$ units left

3. 3 $P$ has a zero at $x = 6$. Where is the corresponding zero of $y = 3P(x)$?

   Show answer
   A vertical stretch leaves zeroes fixed.Still at $x = 6$

4. 4 $P$ has a zero at $x = 6$. Where is the corresponding zero of $y = P(x-2)$?

   Show answer
   The zero moves right by $2$.$x = 8$

5. 5 $P(0) = -4$. Find the $y$-intercept of $y = -P(x) + 1$.

   Show answer
   Reflect the value, then add $1$.$5$