Question

comparing properties of quadratic functions given in different forms
function 1
f(x) = 3x² - 12x + 13
function 2
x | y
-8 | 16
-5 | 7
-2 | 4
1 | 7
4 | 16
7 | 31
(a) what is the vertex of function 1? ( , )
(b) what is the vertex of function 2? ( , )
(c) which function has the smaller minimum value?
function 1
function 2
what is the smaller minimum value?

Explanation:

Step1: Find vertex of Function 1

The function \( f(x) = 3x^2 - 12x + 13 \) is in standard form \( ax^2 + bx + c \). The x - coordinate of the vertex is \( x = -\frac{b}{2a} \). Here, \( a = 3 \), \( b=- 12 \), so \( x=-\frac{-12}{2\times3}=\frac{12}{6} = 2 \). Substitute \( x = 2 \) into \( f(x) \): \( f(2)=3\times(2)^2-12\times2 + 13=3\times4-24 + 13=12-24 + 13=1 \). So vertex of Function 1 is \( (2,1) \).

Step2: Find vertex of Function 2

For a quadratic function (since the table is symmetric around \( x=-2 + 3=1? \) Wait, looking at the table: when \( x=-8 \), \( y = 16 \); \( x=-5 \), \( y = 7 \); \( x=-2 \), \( y = 4 \); \( x = 1 \), \( y = 7 \); \( x = 4 \), \( y = 16 \); \( x = 7 \), \( y = 31 \). The axis of symmetry is the mid - point between \( x=-8 \) and \( x = 4 \) (since \( y \) values are equal at \( x=-8 \) and \( x = 4 \)). The mid - point \( x=\frac{-8 + 4}{2}=\frac{-4}{2}=-2? \) Wait, no, when \( x=-5 \) and \( x = 1 \), \( y = 7 \). Mid - point of \( x=-5 \) and \( x = 1 \) is \( x=\frac{-5 + 1}{2}=\frac{-4}{2}=-2 \). Wait, the minimum \( y \) value is at \( x=-2 \), \( y = 4 \). Wait, no, let's check the symmetry. The function is symmetric about the vertical line through the vertex. The \( x \) values are symmetric around \( x=\frac{-8 + 4}{2}=-2? \) Wait, \( x=-8 \) and \( x = 4 \): difference is 12, mid - point is \( x=-2 \). \( x=-5 \) and \( x = 1 \): difference is 6, mid - point is \( x=-2 \). \( x=-2 \) is the center. So the vertex is at \( x=-2 \), \( y = 4 \)? Wait, no, when \( x=-2 \), \( y = 4 \), and the function is symmetric around \( x=-2 \)? Wait, no, when \( x=-8 \) (distance from \( x=-2 \) is \(|-8+2| = 6\)) and \( x = 4 \) (distance from \( x=-2 \) is \(|4 + 2|=6\)), \( y \) values are both 16. When \( x=-5 \) (distance 3 from \( x=-2 \)) and \( x = 1 \) (distance 3 from \( x=-2 \)), \( y \) values are both 7. So the vertex is at \( x=-2 \), \( y = 4 \)? Wait, no, the vertex is the minimum point. Since the \( y \) values increase as we move away from \( x=-2 \) (from \( x=-2 \), moving left to \( x=-5,-8 \), \( y \) increases from 4 to 7 to 16; moving right to \( x = 1,4,7 \), \( y \) increases from 4 to 7 to 16 to 31). So the vertex of Function 2 is \( (-2,4) \).

Step3: Compare minimum values

The minimum value of a quadratic function \( ax^2+bx + c \) ( \( a>0 \)) is the \( y \) - coordinate of the vertex. For Function 1, minimum value is 1. For Function 2, minimum value is 4. So Function 1 has the smaller minimum value, and the smaller minimum value is 1.

Answer:

s:
(a) The vertex of Function 1 is \((2,1)\)
(b) The vertex of Function 2 is \((-2,4)\)
(c) Function 1 has the smaller minimum value. The smaller minimum value is \(1\)