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17. (4 points) find the inverse of the following function. $y = 4x - 11…

Question

  1. (4 points) find the inverse of the following function.

$y = 4x - 11$

  1. (9 points) given the quadratic function $f(x) = -24x + 9 + 3x^2$ identify the vertex, y - intercept, and if it will have a minimum or maximum value.

(3 pts) standard form:
(3 pts) vertex:
(3 pts) min/max?

  1. (4 points) solve for x.

$(x + 5)^2 - 23 = 98$
$x = $

Explanation:

Response
Question 17: Find the inverse of \( y = 4x - 11 \)

Step 1: Swap \( x \) and \( y \)

To find the inverse, we first swap the roles of \( x \) and \( y \). So we get \( x = 4y - 11 \).

Step 2: Solve for \( y \)

We need to isolate \( y \). First, add 11 to both sides of the equation: \( x + 11 = 4y \). Then, divide both sides by 4: \( y=\frac{x + 11}{4}\). We can also write this as \( y=\frac{1}{4}x+\frac{11}{4}\).

The standard form of a quadratic function is \( f(x)=ax^{2}+bx + c \) (or the vertex form \( f(x)=a(x - h)^{2}+k \), but usually for standard form in terms of \( ax^{2}+bx + c \), we just rearrange the given function. The given function is \( f(x)=3x^{2}-24x + 9 \) (by rearranging the terms in descending order of the power of \( x \)). If we want the vertex form (which is also a type of standard form for vertex - related analysis), we can complete the square.
First, factor out the coefficient of \( x^{2} \) from the first two terms: \( f(x)=3(x^{2}-8x)+9 \).
To complete the square inside the parentheses, we take half of the coefficient of \( x \) (which is \( - 8 \)), square it: \( (\frac{-8}{2})^{2}=16 \).
We add and subtract 16 inside the parentheses: \( f(x)=3(x^{2}-8x + 16-16)+9 \).
Rewrite the expression: \( f(x)=3((x - 4)^{2}-16)+9 \).
Distribute the 3: \( f(x)=3(x - 4)^{2}-48 + 9=3(x - 4)^{2}-39 \).

For a quadratic function in vertex form \( f(x)=a(x - h)^{2}+k \), the vertex is \( (h,k) \). From the vertex form we found above \( f(x)=3(x - 4)^{2}-39 \), we can see that \( h = 4 \) and \( k=-39 \).
If we use the formula for the x - coordinate of the vertex \( h=-\frac{b}{2a} \) for the quadratic function in the form \( f(x)=ax^{2}+bx + c \). For \( f(x)=3x^{2}-24x + 9 \), \( a = 3 \), \( b=-24 \). Then \( h=-\frac{-24}{2\times3}=\frac{24}{6} = 4 \). Then we find \( k=f(4)=3\times4^{2}-24\times4 + 9=3\times16-96 + 9=48-96 + 9=-39 \).

Answer:

The inverse function is \( y=\frac{x + 11}{4} \) (or \( y=\frac{1}{4}x+\frac{11}{4} \))

Question 18: Given the quadratic function \( f(x)=- 24x + 9+3x^{2} \), identify the standard form, vertex, and if it has a minimum or maximum value.
(a) Standard Form