Below are the skills and knowledge that students in the content domain and performance score band selected above are typically able to demonstrate as well as examples of the kinds of questions that these students are likely able to answer correctly. To view skill/knowledge statements and example questions in other domains and/or performance score bands, update the selections above and click Go.

Skills

A student in this performance score band can typically demonstrate the following skills in this content domain:

  • Solve problems with or without context involving one or more nonlinear equations to find the value of an unknown constant
  • Solve a complex equation or formula for a variable of interest; (SAT only) use properties of exponents and properties of polynomial, rational, and radical expressions to rewrite complex expressions, using structure when present, or determine the most suitable form of an equation to display a certain feature

Example Questions

Example Question 1

A quadratic function models the height, in feet, of an object above the ground in terms of the time, in seconds, after the object is launched off an elevated surface. The model indicates the object has an initial height of 10 feet above the ground and reaches its maximum height of 1,034 feet above the ground 8 seconds after being launched. Based on the model, what is the height, in feet, of the object above the ground 10 seconds after being launched?

  1. 234

  2. 778

  3. 970

  4. 1,014

Key: C

Key Explanation

Choice C is correct. It's given that a quadratic function models the height, in feet, of an object above the ground in terms of the time, in seconds, after the object is launched off an elevated surface. This quadratic function can be defined by an equation of the form fx=ax-h2+k, where fx is the height of the object x seconds after it was launched, and a, h, and k are constants such that the function reaches its maximum value, k, when x=h. Since the model indicates the object reaches its maximum height of 1,034 feet above the ground 8 seconds after being launched, fx reaches its maximum value, 1,034, when x=8. Therefore, k=1,034 and h=8. Substituting 8 for h and 1,034 for k in the function fx=ax-h2+k yields fx=ax-82+1,034. Since the model indicates the object has an initial height of 10 feet above the ground, the value of fx is 10 when x=0. Substituting 0 for x and 10 for fx in the equation fx=ax-82+1,034 yields 10=a0-82+1,034, or 10=64a+1,034. Subtracting 1,034 from both sides of this equation yields 64a=-1,024. Dividing both sides of this equation by 64 yields a=-16. Therefore, the model can be represented by the equation fx=-16x-82+1,034. Substituting 10 for x in this equation yields f10=-1610-82+1,034, or f10=970. Therefore, based on the model, 10 seconds after being launched, the height of the object above the ground is 970 feet.

Distractor Explanations

Choice A is incorrect and may result from conceptual or calculation errors.

Choice B is incorrect and may result from conceptual or calculation errors.

Choice D is incorrect and may result from conceptual or calculation errors.

Example Question 2

Function f is defined by fx=-ax+b, where a and b are constants. In the xy-plane, the graph of y=fx-12 has a y-intercept at 0,-757. The product of a and b is 320 7 . What is the value of a ?

Key: 20

Key Explanation

The correct answer is 20. It’s given that fx=-ax+b. Substituting -ax+b for fx in the equation y=fx-12 yields y=-ax+b-12. It’s given that the y-intercept of the graph of y=fx-12 is 0,-757. Substituting 0 for x and -757 for y in the equation y=-ax+b-12 yields -757=-a0+b-12, which is equivalent to -757=-1+b-12, or -757=b-13. Adding 13 to both sides of this equation yields 167=b. It’s given that the product of a and b is 3207, or ab=3207. Substituting 167 for b in this equation yields a167=3207. Dividing both sides of this equation by 167 yields a=20.