The solution:
To find the Axis of symmetry you need to solve for the Vertex (h, k)(h,k)
Formula for the vertex:
h=(-b)/(2a)h=−b2a and k=c-b^2/(4a)k=c−b24a
From the given y=2x^2-4x-3y=2x2−4x−3
a=2a=2 and b=-4b=−4 and c=-3c=−3
h=(-b)/(2a)=(-(-4))/(2(2))=1h=−b2a=−(−4)2(2)=1
k=c-b^2/(4a)=-3-(-4)^2/(4(2))=-5k=c−b24a=−3−(−4)24(2)=−5
Axis of symmetry:
x=hx=h
color(blue)(x=1)x=1
Since aa is positive, the function has a Minimum value and does not have a Maximum.
Minimum value color(blue)(=k=-5)=k=−5
The graph of y=2x^2-4x-3y=2x2−4x−3
Desmos.com
To draw the graph of y=2x^2-4x-3y=2x2−4x−3, use the vertex (h, k)=(1, -5)(h,k)=(1,−5) and the intercepts.
When x=0x=0,
y=2x^2-4x-3y=2x2−4x−3
y=2(0)^2-4(0)-3=-3" "y=2(0)2−4(0)−3=−3 means there is a point at (0, -3)(0,−3)
and when y=0y=0,
y=2x^2-4x-3y=2x2−4x−3
0=2x^2-4x-30=2x2−4x−3
x=(-b+-sqrt(b^2-4ac))/(2a)=(-(-4)+-sqrt((-4)^2-4(2)(-3)))/(2(2))x=−b±√b2−4ac2a=−(−4)±√(−4)2−4(2)(−3)2(2)
x=(+4+-sqrt(16+24))/(4)x=+4±√16+244
x=(+4+-sqrt(40))/(4)x=+4±√404
x=(+4+-2sqrt(10))/(4)x=+4±2√104
x_1=1+1/2sqrt(10)x1=1+12√10
x_2=1-1/2sqrt(10)x2=1−12√10
We have two points at (1+1/2sqrt(10), 0)(1+12√10,0) and (1-1/2sqrt(10), 0)(1−12√10,0)
God bless...I hope the explanation is useful.
