y=(x+4)^{2}-7 a) Determine the x-intercepts of the quadratic equation in both

Step 1 
Solution: ${\displaystyle y={(x+4)}^2-7}$ 
${\displaystyle {(x+4)}^2=y+7}$ 
1) By substituting ${\displaystyle y=0}$, the x intercept can be achieved.
${\displaystyle {(x+4)}^2=0+7}$ 
${\displaystyle {(x+4)}^2=7}$ 
${\displaystyle x+4=\pm \sqrt{7}}$ 
${\displaystyle x=\pm \sqrt{7}-4}$ 
Because of this, x intercepts are  ${\displaystyle \sqrt{7}-4}$ and ${\displaystyle -\sqrt{7}-4}$ 
2) contrasting it with the fundamental parabola equation
${\displaystyle {(x-h)}^2=4a(y-k)}$ 
${\displaystyle {(x+4)}^2=(y+7)}$ 
${\displaystyle h=-4,k=-7}$ 
Therefore vertex is ${\displaystyle (h,k)=(-4,-7)}$ 
Step 2 
3) You can get the y intercept by changing ${\displaystyle x=0}$ 
${\displaystyle y={(x+4)}^2-7}$ 
${\displaystyle y={(0+4)}^2-7}$ 
${\displaystyle y=16-7=9}$ 
4) ${\displaystyle {(x-h)}^2=4a(y-k)}$ 
${\displaystyle {(x+4)}^2=(y+7)}$ 
${\displaystyle \therefore 4a=1}$ 
${\displaystyle a=\frac{1}{4}}$ 
Therefore four of the parabola is of ${\displaystyle (0,y_4)}$ 
Also ${\displaystyle y={(x+4)}^2-7}$ 
${\displaystyle y=x^2+16+8x-7}$ 
${\displaystyle y=x^2+8x+9}$ 
Coefficient of ${\displaystyle x^2=1\Rightarrow \text{Parabola}}$ opens up wards. 

Step 1
a) Write the equation in standard form
$-x^2-8x+y-9=0$
Step 2
Substitute the values of a, b, and c into the quadratic formula
$x=\frac{-(-8)-\sqrt{(-8{)}^2-4(-1)(y-9)}}{2(-1)}$ or $x=\frac{-(-8)+\sqrt{(-8{)}^2-4(-1)(y-9)}}{2(-1)}$
Step 3
Simplify the discriminant
$x=\frac{-(8)-\sqrt{4(y+7)}}{2(-1)}orx=\frac{-(-8)+\sqrt{4(y+7)}}{2(-1)}$
Step 4
Simplify
$x=\sqrt{y+7}-4orx=-\sqrt{y+7}-4$
Step 5
b) Let's focus on:
$(x+4{)}^2-7$
Expand the expression
$x^2+8x+16-7$
Step 6
Group like terms together
$x^2+8x+(16-7)$
Step 7
Add or subtract the numbers
$x^2+8x+9$
c) Parabola equation in polynomial form
The vertex ofan up-down facing parabola of the form $y=a{x}^2+bx+c$ is $x_v=-\frac{b}{2a}$
Rewrite $y=(x+4{)}^2-7$ in the form $y=a{x}^2+bx+c$
$y=x^2+8x+9$
The parabola params are:
$a=1,b=8,x=9$
$x_v=-\frac{b}{2a}$
$x_v=-\frac{8}{2\times 1}$
Simplify
$x_v=-4$
Plug in $x_v=-4$ to find the $y_v$ value
$y_v=-7$
Therefore the parabola vertex is
(-4, -7)
If a<0, then the vertex is a maximum value
If Pa>0, then the vertex is a minimum value
Minimum (-4, -7)
Step 8