Question

PLEASE HELP

Answer 1

Answer:

Part 1) The minimum value is $(5,-9)$

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 $(-3,2)$

Step-by-step explanation:

Part 1) we have

$g(x)=x^{2}-10x+16$

we know that

The equation of a vertical parabola in vertex form is equal to

$y=a(x-h)^{2}+k$

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

$g(x)-16=x^{2}-10x$

$g(x)-16+25=x^{2}-10x+25$

$g(x)+9=x^{2}-10x+25$

$g(x)+9=(x-5)^{2}$

$g(x)=(x-5)^{2}-9$ ------> vertex form

The vertex is the point $(5,-9)$

the parabola open upward (vertex is a minimum)

Part 2) we have

$f(x)=x^{2}-3x-10$

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

$x^{2}-3x-10=0$

The formula to solve a quadratic equation of the form $ax^{2} +bx+c=0$ is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$x^{2}-3x-10=0$

so

$a=1\\b=-3\\c=-10$

substitute in the formula

$x=\frac{-(-3)(+/-)\sqrt{-3^{2}-4(1)(-10)}} {2(1)}$

$x=\frac{3(+/-)\sqrt{49}} {2}$

$x=\frac{3(+/-)7} {2}$

$x=\frac{3(+)7} {2}=5$

$x=\frac{3(-)7} {2}=-2$

Part 3) we have

$f(x)=x^{2}+17x+72$

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

$x^{2}+17x+72=0$

The formula to solve a quadratic equation of the form $ax^{2} +bx+c=0$ is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$x^{2}+17x+72=0$

so

$a=1\\b=17\\c=72$

substitute in the formula

$x=\frac{-17(+/-)\sqrt{17^{2}-4(1)(72)}} {2(1)}$

$x=\frac{-17(+/-)\sqrt{1}} {2}$

$x=\frac{-17(+/-)1} {2}$

$x=\frac{-17(+)1} {2}=-8$

$x=\frac{-17(-)1} {2}=-9$

Part 4) we have

$f(x)=x^{2}+6x+11$

we know that

The equation of a vertical parabola in vertex form is equal to

$y=a(x-h)^{2}+k$

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

$f(x)-11=x^{2}+6x$

$f(x)-11+9=x^{2}+6x+9$

$f(x)-2=x^{2}+6x+9$

$f(x)-2=(x+3)^{2}$

$f(x)=(x+3)^{2}+2$

The vertex is the point $(-3,2)$

the parabola open upward (vertex is a minimum)