Respuesta :
vertex for function 1 : (-2,4)
vertex for function 2: (2,7)
function 1 has the smaller value
it is 4
Explanation
given
[tex]f(x)=ax^2+bx+c[/tex]to get the vertex from the quadratic function we need to use
[tex]\begin{gathered} h=-\frac{b}{2a} \\ so \\ \text{vertex} \\ (h,f(h)) \end{gathered}[/tex]then
Let
Step 1
Function 2
Let
[tex]\begin{gathered} f(x)=3x^2-12x+19\Rightarrow ax^2+bx+c \\ \text{hence. a=3 b=-12} \\ h=-\frac{b}{2a} \\ h=-\frac{-12}{2(3)}=\frac{12}{6}=2 \\ \text{now, f(h), so} \\ f(2)=3(2)^2-12(2)+19 \\ f(2)=12-24+19=31-24=7 \end{gathered}[/tex]hence, the vertex is
vertex: (2,7)
Step 2
function 1
we have the table, so
a) find the function
so
for x= 1
[tex]\begin{gathered} f(x)=ax^2+bx+c \\ f(1)=a(1)^2+b(1)+c \\ f(1)=a+b+c \\ \text{ in the table we have ( 1,7) so} \\ 7=a+b+c\Rightarrow equation(1) \end{gathered}[/tex]for x= -2
[tex]\begin{gathered} f(x)=ax^2+bx+c \\ f(-2)=a(-2)^2+b(-2)+c \\ f(-2)=4a-2b+c \\ \text{ in the table we have ( -2,4) so} \\ 4=4a-2b+c\Rightarrow equation(2) \end{gathered}[/tex]for x= 4
[tex]\begin{gathered} f(x)=ax^2+bx+c \\ f(4)=a(4)^2+b(4)+c \\ f(4)=16a+4b+c \\ \text{ in the table we have ( 4,16) so} \\ 16=16a+4b+c\Rightarrow equation(3) \end{gathered}[/tex]now, we have a system of 3 variables and 3 equations, so
[tex]\begin{gathered} 7=a+b+c\Rightarrow equation(1) \\ 4=4a-2b+c\Rightarrow equation(2) \\ 16=16a+4b+c\Rightarrow equation(3) \end{gathered}[/tex]a) subtract equation (2) from equation (1) to eliminate c
[tex]\begin{gathered} 7=a+b+c\Rightarrow equation(1) \\ -(4=4a-2b+c)\Rightarrow equation(2) \\ _{--------------} \\ 3=-3a+3b\Rightarrow equation(4) \end{gathered}[/tex]b) subtract eq(3) from eq(1) to eliminate c
[tex]\begin{gathered} 7=a+b+c\Rightarrow equation(1) \\ -(16=16a+4b+c) \\ _{--------------} \\ -9=-15a-3b\Rightarrow equation\text{ (5)} \end{gathered}[/tex]c) now, add equation (4) and equation(5) to eliminate b
[tex]\begin{gathered} 3=-3a+3b\Rightarrow equation(4) \\ -9=-15a-3b\Rightarrow equation\text{ (5)} \\ ----------- \\ -6=-18a \\ \text{divide both sides by -18} \\ \frac{-6}{-18}=\frac{-18a}{-18} \\ \frac{1}{3}=a \end{gathered}[/tex]now, replace in equation (4) to find b
[tex]\begin{gathered} 3=-3a+3b\Rightarrow equation(4) \\ 3=-3(\frac{1}{3})+3b\Rightarrow equation(4) \\ 3=-1+3b \\ \text{add}1\text{ in both sides} \\ 3+1=-1+3b+1 \\ 4=3b \\ \text{divide both sides by 3} \\ \frac{4}{3}=b \end{gathered}[/tex]d) finally, replace a and b values into equation (1) , then solve for c
[tex]\begin{gathered} 7=a+b+c\Rightarrow equation(1) \\ 7=\frac{1}{3}+\frac{4}{3}+c\Rightarrow equation(1) \\ 7=\frac{5}{3}+c \\ \text{subtract 5/3 in both sides} \\ \frac{21}{3}-\frac{5}{3}=c \\ \frac{16}{3}=C \end{gathered}[/tex]hence, the function is
[tex]\begin{gathered} f(x)=ax^2+bx+c \\ f(x)=\frac{1}{3}x^2+\frac{4}{3}x+\frac{16}{3} \end{gathered}[/tex]now, apply the formula to find the vertex
[tex]\begin{gathered} f(x)=\frac{1}{3}x^2+\frac{4}{3}x+\frac{16}{3}\Rightarrow ax^2+bx+c \\ \text{hence. a=1/3 b=}\frac{\text{4}}{3} \\ h=-\frac{\frac{4}{3}}{2(\frac{1}{3})}=\frac{\frac{4}{3}}{\frac{2}{3}}=-\frac{12}{6}=-2 \\ h=-2 \\ \text{now, f(h), so} \\ f(-2)=\frac{1}{3}(-2)^2+\frac{4}{3}(-2)+\frac{16}{3} \\ f(-2)=\frac{4}{3}-\frac{8}{3}+\frac{16}{3}=\frac{12}{3}=4 \\ so,\text{ the vertex is } \\ (-2,4) \end{gathered}[/tex]vertex for function 1 : (-2,4)
Step 3
smaller minimum value:
let's compare the image of the vertex
vertex for function 1 : (-2,4)
vertex for function 2: (2,7)
the y coordinate the function 2 is greater, so
function 1 has the smaller value
it is 4
I hope this helps you
