Quadratic Graphs

Question 1

Which of the following graphs represent $y = x2- 4 ?$

Not correct. Choice (a) is false.

Try again, this is the line $y = x- 4.$

Not correct. Choice (b) is false.

Try again, this is the graph of the equation $2 y = x - x- 4.$

Your answer is correct.

$y = x2-4$ is concave up and it has a turning point at $(0,-4).$

Not correct. Choice (d) is false.

Try again, this graph $2 y = 4- x$ which is concave down.

Question 2

Consider the graph of $2 y = ax$ below.

Which of the following statements about the graph is true?

Your answer is correct.

The graph is concave down so it has a maximum and $a$ is negative.

Not correct. Choice (b) is false.

Try again, the graph has a maximum but you have the coefficient right.

Not correct. Choice (c) is false.

Try again, if the graph has a maximum then $a$ must be negative.

Not correct. Choice (d) is false.

The graph is concave down so it has a maximum.

Question 3

Which of the following graphs represents $y = b0 + b1x + b2x2$ where $b0 = 1 and b2 < 0?$

1.		2.

3.		4.

Not correct. Choice (a) is false.

Try again, 1 has $b < 0 2$ but 2 has $b > 0 . 2$

Not correct. Choice (b) is false.

Try again, 3 has $b < 0 2$ but 2 has $b > 0. 2$

Your answer is correct.

All of the graphs have $b = 1 0$ but only 1 and 3 are concave down, as required.

Not correct. Choice (d) is false.

Try again, both have $b > 0. 2$

Question 4

A parabola with equation $f(x) = b0 + b1x+ b2x2$ has a turning point at $b1- x = - 2b2 .$ The concavity of the parabola determines whether it is a maximum or a minimum.

Consider $f(x) = - 4+ 3x + x2.$ Which one of the following statements is true?

Not correct. Choice (a) is false.

Try again, if $b > 0 2$ then $f(x)$ has a minimum.

Your answer is correct.

Since $b2 > 0, f(x)$ has a minimum at $x = - 3. 2$ and $( )2 f(-3 ) = -4 + 3× - 3+ - 3 = - 4- 9 + 9 = -61 . 2 2 2 2 4 4$

Not correct. Choice (c) is false.

Try again, you have not substituted into the equation correctly.

Not correct. Choice (d) is false.

Try again, if $b2 > 0$ then $f (x)$ has a minimum and you have not substituted into the equation correctly.

Question 5

Which of the parabolic functions below have $y$ -intercept of 8 , an $x$ -intercept of 2 and a maximum turning point?

Not correct. Choice (a) is false.

Try again, $f(0) = 8$ and $f(2) = 0$ but the parabola is concave up and hence has a minimum turning point.

Not correct. Choice (b) is false.

Try again, $f(0) = 8$ and the coefficient of $x2$ is negative so we have a maximum but
$f(2) = 8 + 4- 4 = 8$ so the $x$ -intercept is not 2 .

Not correct. Choice (c) is false.

Try again, the $y$ -intercept is -8 . The other requirements are met.

Your answer is correct.

Since $f(0) = 8, f (2) = 8- 4- 4 = 0$ and the coefficient of $x2$ is negative we have satisfied all of the requirements.


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