Question

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using tables on your calculator
common core algebra ii
the graphing calculator is an amazing device that can do many things. one function that it is particularly good at is evaluating expressions for different input values. we will be looking at two tools on the calculator today, the store feature and tables. first lets look at how to use store.
exercise #1: find the value of each of the following expressions by using the store feature on your calculator.
(a) $x^2 - 2x + 7$ for $x = 5$
(b) $\frac{2x + 6}{3x - 5}$ for $x = -10$
(c) $|7x - 20| + x^3$ for $x = 2$

sometimes the calculator can even tell us useful information even when it has a hard time evaluating an expression.
exercise #2: consider the expression $sqrt{6 - 2x}$. what happens when you try to use store to evaluate this expression for $x = 5$? evaluate the expression by hand to help explain what the calculator is trying to tell us.

exercise #3: lets work with the product of two binomials again, specifically $(3x + 2)$ and $(x + 5)$.
(a) find their product in trinomial form.
(b) evaluate both the trinomial and the original product for $x = -2$. what do you notice?
(c) use the store command to evaluate the trinomial from (a) for $x = -5$. why does the value of the trinomial turn out to be this specific value at $x = -5$? explain.

common core algebra ii, unit #1 - essential algebra concepts - lesson #6

Explanation:

Exercise #1

(a) Step1: Substitute x=5 into expression

$5^2 - 2(5) + 7$

(a) Step2: Calculate each term

$25 - 10 + 7$

(a) Step3: Simplify the expression

$25-10=15; 15+7=22$

(b) Step1: Substitute x=-10 into expression

$\frac{2(-10)+6}{3(-10)-5}$

(b) Step2: Calculate numerator and denominator

$\frac{-20+6}{-30-5}=\frac{-14}{-35}$

(b) Step3: Simplify the fraction

$\frac{-14}{-35}=\frac{2}{5}$

(c) Step1: Substitute x=2 into expression

$|7(2)-20| + 2^2$

(c) Step2: Calculate each term

$|14-20| + 4 = |-6| + 4$

(c) Step3: Simplify the expression

$6 + 4 = 10$

Exercise #2

Step1: Substitute x=5 into expression

$\sqrt{6-2(5)}=\sqrt{6-10}=\sqrt{-4}$

Step2: Analyze the result

Square roots of negative numbers are not real numbers, so the calculator will return an error (or indicate a non-real result, as real-number calculators cannot compute this).

Exercise #3

(a) Step1: Expand the binomial product

$(3x+2)(x+5)=3x(x+5)+2(x+5)$

(a) Step2: Distribute terms

$3x^2 + 15x + 2x + 10$

(a) Step3: Combine like terms

$3x^2 + 17x + 10$

(b) Step1: Evaluate original product at x=-2

$(3(-2)+2)((-2)+5)=(-6+2)(3)=(-4)(3)=-12$

(b) Step2: Evaluate trinomial at x=-2

$3(-2)^2 +17(-2)+10=3(4)-34+10=12-34+10=-12$

(b) Step3: Compare results

Both expressions have the same value.

(c) Step1: Evaluate trinomial at x=-5

$3(-5)^2 +17(-5)+10=3(25)-85+10=75-85+10=0$

(c) Step2: Explain the result

$x=-5$ is a root of the binomial $(x+5)$, so the original product $(3x+2)(x+5)$ equals 0 when $x=-5$, and the trinomial (its expanded form) also equals 0.

Answer:

Exercise #1

(a) 22
(b) $\frac{2}{5}$
(c) 10

Exercise #2

The calculator will return an error (or indicate a non-real result). By hand, $\sqrt{6-2(5)}=\sqrt{-4}$, which is not a real number, so the calculator cannot compute a real output for this input.

Exercise #3

(a) $3x^2 + 17x + 10$
(b) Both the original product and the trinomial equal -12; they are equivalent expressions, so they have the same value for any x.
(c) The value is 0. $x=-5$ makes the factor $(x+5)=0$, so the product (and its expanded trinomial form) equals 0.