Question Which of these statements is true about the quadratic functions $f(x)=-15 x^{2}+32$ and $g(x)=-17 x^{2}+5 x+32$ ?

Studdy Solution

STEP 1

Assumptions
1. The functions given are $f(x)=-15x^{2}+32$ and $g(x)=-17x^{2}+5x+32$ .
2. We are comparing the zeros, maximum values, $y$ -intercepts, and the sharpness of the rise and fall of the two functions.
3. Both functions are quadratic, and their standard form is $ax^{2}+bx+c$ .

STEP 2

To determine if $f(x)$ and $g(x)$ have the same zeros, we need to find the roots of each function by setting them equal to zero and solving for $x$ .
$f(x) = 0 \Rightarrow -15x^{2}+32 = 0$ $g(x) = 0 \Rightarrow -17x^{2}+5x+32 = 0$

STEP 3

Solve the equation $-15x^{2}+32 = 0$ for $x$ to find the zeros of $f(x)$ .
$-15x^{2} = -32$ $x^{2} = \frac{32}{15}$ $x = \pm\sqrt{\frac{32}{15}}$

STEP 4

Solve the equation $-17x^{2}+5x+32 = 0$ for $x$ to find the zeros of $g(x)$ . This requires using the quadratic formula, $x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ , where $a = -17$ , $b = 5$ , and $c = 32$ .
$x = \frac{-5 \pm \sqrt{5^{2}-4(-17)(32)}}{2(-17)}$

STEP 5

Calculate the discriminant for $g(x)$ , which is $b^{2}-4ac$ .
$\Delta = 5^{2}-4(-17)(32)$ $\Delta = 25 + 4(17)(32)$ $\Delta = 25 + 2176$ $\Delta = 2201$

STEP 6

Since the discriminant for $g(x)$ is positive, there are two distinct real roots. Calculate the roots using the quadratic formula.
$x = \frac{-5 \pm \sqrt{2201}}{2(-17)}$ $x = \frac{-5 \pm 47}{-34}$

STEP 7

Find the two roots for $g(x)$ .
$x_1 = \frac{-5 + 47}{-34}$ $x_2 = \frac{-5 - 47}{-34}$

STEP 8

Simplify the roots for $g(x)$ .
$x_1 = \frac{42}{-34} = -\frac{21}{17}$ $x_2 = \frac{-52}{-34} = \frac{26}{17}$

STEP 9

Compare the zeros of $f(x)$ and $g(x)$ . Since the zeros of $f(x)$ are $\pm\sqrt{\frac{32}{15}}$ and the zeros of $g(x)$ are $-\frac{21}{17}$ and $\frac{26}{17}$ , they are not the same.

STEP 10

To determine if $f(x)$ has a higher maximum than $g(x)$ , we need to compare the $y$ -coordinates of their vertices, since both functions open downwards (the coefficients of $x^2$ are negative).

STEP 11

The vertex of a parabola given by $ax^{2}+bx+c$ is at the point $(-\frac{b}{2a}, f(-\frac{b}{2a}))$ . Calculate the vertex of $f(x)$ .
$x_{vertex} = -\frac{b}{2a} = -\frac{0}{2(-15)} = 0$ $y_{vertex} = f(0) = -15(0)^{2}+32 = 32$

STEP 12

Calculate the vertex of $g(x)$ .
$x_{vertex} = -\frac{b}{2a} = -\frac{5}{2(-17)} = \frac{5}{34}$ $y_{vertex} = g\left(\frac{5}{34}\right) = -17\left(\frac{5}{34}\right)^{2}+5\left(\frac{5}{34}\right)+32$

STEP 13

Simplify the $y$ -coordinate of the vertex of $g(x)$ .
$y_{vertex} = -17\left(\frac{25}{1156}\right)+\frac{25}{34}+32$ $y_{vertex} = -\frac{425}{1156}+\frac{850}{1156}+32$ $y_{vertex} = \frac{425}{1156}+32$

STEP 14

Since $32 > \frac{425}{1156}$ , we can conclude that $f(x)$ has a higher maximum than $g(x)$ .

STEP 15

To determine if $f(x)$ and $g(x)$ have the same $y$ -intercept, we need to evaluate both functions at $x=0$ .
$f(0) = -15(0)^{2}+32 = 32$ $g(0) = -17(0)^{2}+5(0)+32 = 32$

STEP 16

Since $f(0) = g(0) = 32$ , we can conclude that $f(x)$ and $g(x)$ have the same $y$ -intercept.

STEP 17

To determine if $f(x)$ rises and falls more sharply than $g(x)$ , we need to compare the absolute values of the coefficients of $x^2$ in both functions.

STEP 18

Compare the coefficients of $x^2$ in $f(x)$ and $g(x)$ .
$|a_{f(x)}| = |-15| = 15$ $|a_{g(x)}| = |-17| = 17$

STEP 19

Since $|a_{g(x)}| > |a_{f(x)}|$ , we can conclude that $g(x)$ rises and falls more sharply than $f(x)$ .

STEP 20

Review the statements given in the problem and match them with the conclusions we have drawn:
1. $f(x)$ and $g(x)$ have the same zeros. (False, as shown in STEP_9)
2. $f(x)$ has a higher maximum than $g(x)$ . (True, as shown in STEP_14)
3. $f(x)$ and $g(x)$ have the same $y$ -intercept. (True, as shown in STEP_16)
4. $f(x)$ rises and falls more sharply than $g(x)$ . (False, as shown in STEP_19)

STEP 21

Based on the analysis, the true statement about the functions $f(x)$ and $g(x)$ is:
$f(x)$ and $g(x)$ have the same $y$ -intercept.

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