Question 92516
1.Find the y-intercept of y=2^x.
The y intercept occurs when x = 0, we know that any number to the power of 0 is equal to 1.
y = 2^0
y = 1
:
2. Use quadratic formula to solve:5x^2+12=-6x. Put into the ax^2 + bx + c = 0 format.
5x^2 + 12 = -6x
5x^2 + 6x + 12 = 0
:
a = 5; b = 6; c = 12
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
:
{{{x = (-6 +- sqrt( 6^2 - 4 * 5 * 12 ))/(2*5) }}}
:
{{{x = (-6 +- sqrt( 36 - 108 ))/(10) }}}
:
You can see that the discriminant will be < 0, no real roots
:
: 
3. Use the quadratic formula to solve:x^2+12x+11=0.
a = 1; b = 12; c = 11
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
:
{{{x = (-12 +- sqrt( 12^2 - 4 * 1 * 11 ))/(2*1) }}}
:
{{{x = (-12 +- sqrt( 144 - 44 ))/(2) }}}
:
{{{x = (-12 +- sqrt(100 ))/(2) }}}
:
Two solutions:
{{{x = (-12 + 10)/(2) }}}
{{{x = -2/2 }}}
x = -1
and
{{{x = (-12 -10)/(2) }}}
{{{x = -22/2
x = -11
:
:
4.Describe how the graph of y=(x-6)^2+5 changes from the parent graph of y=x^2.
Minimum is at origin in y = x^2, minimum is at x/y coordinate 6,5 for y = (x-6)^2 + 5
: