Answer:
Part 1) The minimum value is [tex](5,-9)[/tex]
Part 2) The x-intercepts are -2 and 5
Part 3) The zeros of the function are -9 and -8
Part 4) The minimum value is [tex](-3,2)[/tex]
Step-by-step explanation:
Part 1) we have
[tex]g(x)=x^{2}-10x+16[/tex]
we know that
The equation of a vertical parabola in vertex form is equal to
[tex]y=a(x-h)^{2}+k[/tex]
where
(h,k) is the vertex
if a>0---> the the parabola open upward (vertex is a minimum)
If a<0---> the the parabola open downward (vertex is a maximum)
Convert to vertex form
[tex]g(x)-16=x^{2}-10x[/tex]
[tex]g(x)-16+25=x^{2}-10x+25[/tex]
[tex]g(x)+9=x^{2}-10x+25[/tex]
[tex]g(x)+9=(x-5)^{2}[/tex]
[tex]g(x)=(x-5)^{2}-9[/tex] ------> vertex form
The vertex is the point [tex](5,-9)[/tex]
the parabola open upward (vertex is a minimum)
Part 2) we have
[tex]f(x)=x^{2}-3x-10[/tex]
we know that
The x-intercepts are the values of x when the value of the function is equal to zero
so
equate the equation to zero
[tex]x^{2}-3x-10=0[/tex]
The formula to solve a quadratic equation of the form [tex]ax^{2} +bx+c=0[/tex] is equal to
[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]x^{2}-3x-10=0[/tex]
so
[tex]a=1\\b=-3\\c=-10[/tex]
substitute in the formula
[tex]x=\frac{-(-3)(+/-)\sqrt{-3^{2}-4(1)(-10)}} {2(1)}[/tex]
[tex]x=\frac{3(+/-)\sqrt{49}} {2}[/tex]
[tex]x=\frac{3(+/-)7} {2}[/tex]
[tex]x=\frac{3(+)7} {2}=5[/tex]
[tex]x=\frac{3(-)7} {2}=-2[/tex]
Part 3) we have
[tex]f(x)=x^{2}+17x+72[/tex]
we know that
The zeros of the function are the values of x when the value of the function is equal to zero
so
equate the equation to zero
[tex]x^{2}+17x+72=0[/tex]
The formula to solve a quadratic equation of the form [tex]ax^{2} +bx+c=0[/tex] is equal to
[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]x^{2}+17x+72=0[/tex]
so
[tex]a=1\\b=17\\c=72[/tex]
substitute in the formula
[tex]x=\frac{-17(+/-)\sqrt{17^{2}-4(1)(72)}} {2(1)}[/tex]
[tex]x=\frac{-17(+/-)\sqrt{1}} {2}[/tex]
[tex]x=\frac{-17(+/-)1} {2}[/tex]
[tex]x=\frac{-17(+)1} {2}=-8[/tex]
[tex]x=\frac{-17(-)1} {2}=-9[/tex]
Part 4) we have
[tex]f(x)=x^{2}+6x+11[/tex]
we know that
The equation of a vertical parabola in vertex form is equal to
[tex]y=a(x-h)^{2}+k[/tex]
where
(h,k) is the vertex
if a>0---> the the parabola open upward (vertex is a minimum)
If a<0---> the the parabola open downward (vertex is a maximum)
Convert to vertex form
[tex]f(x)-11=x^{2}+6x[/tex]
[tex]f(x)-11+9=x^{2}+6x+9[/tex]
[tex]f(x)-2=x^{2}+6x+9[/tex]
[tex]f(x)-2=(x+3)^{2}[/tex]
[tex]f(x)=(x+3)^{2}+2[/tex]
The vertex is the point [tex](-3,2)[/tex]
the parabola open upward (vertex is a minimum)
