Skill: Graph exponential functions using transformations

Explanation and Free Practice Resources

The learner interprets f(x)=ab x−h+kf(x)=a b^{\,x-h}+k, with b>0b>0 and b≠1b\ne1, as a transformed parent exponential bxb^x: hh shifts horizontally, kk shifts vertically, and aa produces vertical stretch or compression and reflection when negative. Graphs reflect the corresponding growth or decay, horizontal asymptote y=ky=k, domain, range, and key intercept or reference points; the horizontal shift is x−hx-h, not x+hx+h. This treatment focuses on standard real-valued transformations and does not include more advanced parameter analysis or calculus-based behavior.

Detailed Explanation: Graph exponential functions using transformations

To graph an exponential function in the form

f(x)=abx−h+k,f(x)=a b^{x-h}+k,

compare it with the parent function y=bxy=b^x.

  • hh shifts the graph horizontally:
    • x−hx-h means shift right hh units.
    • For example, x−1x-1 means right 11 unit.
  • kk shifts the graph vertically:
    • +k+k means shift up kk units.
  • aa changes the vertical shape:
    • ∣a∣>1 \vert a \vert >1 gives a vertical stretch.
    • 0<∣a∣<10< \vert a \vert <1 gives a vertical compression.
    • If a<0a<0, reflect the graph across the xx-axis.
  • The horizontal asymptote is y=ky=k.
  • The domain of an exponential function is all real numbers.

Example

Graph

f(x)=−2⋅3x−1+4.f(x)=-2\cdot 3^{x-1}+4.

Step 1: Identify the parent function

The parent function is

y=3x.y=3^x.

Since 3>13>1, the parent graph represents exponential growth.

Step 2: Identify the transformations

Rewrite the function in the form abx−h+ka b^{x-h}+k:

f(x)=−2⏟a⋅3x−1⏟bx−h+4⏟k.f(x)=\underbrace{-2}_{a}\cdot \underbrace{3^{x-1}}_{b^{x-h}}+\underbrace{4}_{k}.

Therefore:

  • x−1x-1 shifts the graph right 1 unit.
  • The factor −2-2 reflects the graph across the xx-axis and stretches it vertically by a factor of 22.
  • +4+4 shifts the graph up 4 units.

Notice that x−hx-h determines the shift. Thus, x−1x-1 means right 11, not left 11.

Step 3: Find the horizontal asymptote

The parent function y=3xy=3^x has asymptote y=0y=0.

Shifting up 44 units gives the new horizontal asymptote:

y=4.\boxed{y=4}.

Step 4: Transform some key points

Useful points on y=3xy=3^x are

(−1,13),(0,1),(1,3).(-1,\tfrac13),\qquad (0,1),\qquad (1,3).

For the new function, use

xnew=xparent+1x_{\text{new}}=x_{\text{parent}}+1

because the graph shifts right 11, and

ynew=−2yparent+4.y_{\text{new}}=-2y_{\text{parent}}+4.

Transform the points:

Parent pointNew point(−1,13)(0,103)(0,1)(1,2)(1,3)(2,−2)\begin{array}{c|c} \text{Parent point} & \text{New point} \\ \hline (-1,\tfrac13) & (0,\tfrac{10}{3})\\ (0,1) & (1,2)\\ (1,3) & (2,-2) \end{array}

Plot these points and draw a smooth exponential curve approaching, but never touching, y=4y=4.

Because of the negative coefficient, the graph decreases from left to right.

Step 5: State the important features

For

f(x)=−2⋅3x−1+4:f(x)=-2\cdot 3^{x-1}+4:
  • Horizontal asymptote: y=4\boxed{y=4}

  • Domain: (−∞,∞)\boxed{(-\infty,\infty)}

  • Range: (−∞,4)\boxed{(-\infty,4)}

  • yy-intercept:

    Using the point table, when x=0x=0,

f(0)=103. f(0)=\frac{10}{3}.

So the yy-intercept is (0,103)\boxed{(0,\frac{10}{3})}.

  • xx-intercept:

    Set f(x)=0f(x)=0:

−2⋅3x−1+4=0 -2\cdot 3^{x-1}+4=0 3x−1=2 3^{x-1}=2 x=1+log⁡32. x=1+\log_3 2.

Thus the xx-intercept is approximately

(1.63,0). \boxed{(1.63,0)}.

The final graph is a decreasing exponential curve with asymptote y=4y=4, passing through points such as (0,103)(0,\frac{10}{3}), (1,2)(1,2), and (2,−2)(2,-2).

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