Here is the whole question... I have figured most of it out but I need help with part (b)(f) and (g)... Maybe you also check the work I have already done
Given the following function, find:
(a) vertex, (b) axis of symmetry, (c) intercepts, (d) domain, (e) range,
(f) intervals where the function is increasing,
(g) intervals where the function is decreasing, and
f(x)=-x^2+2x+3
here's my work
y=-x^2+2x+3
y=-1(x^2-2x+1)-1(-3)-(-1)(0+1)
y=1(x^2-2x+1)-1(-3)-(-1)(1)
y=-1(x-1)^2-1(-3)-(-1)(1)
y=-1(x-1)^2+3-(-1)(1)
y=-1(x-1)^2+3+1
y=-1(x-1)^2+4
y=a(x-h)^2+k
a=-1 k=4 h=1
Vetex=(1,4)
axis of symmetry=
(0)=-x^2+2x+3
-x^2+2x+3=0
x^2-2x-3=0
(x+1)(x-3)=0
x=-1,3
y=-(0)^2+2(0)+3
y=-(0)+2(0)+3
y=0+0+3
y=3
-x^2+2x-f(x)=0
-x^2*-1+2x*-1+3*-f(x)*-1=0*-1
x^2-2x+f(x)-3=0*-1
x^2-2x+f(x)-3=0
a=1 b=-2 c= 1f(x)-3
x=-(2)¡À¡Ì((-2)^2-4(1)(1fx)-3)/(2(1))
x=2¡À2 ¡Ì(4-f(x))/2
x=1+¡Ì(4-f(x)
x=2-2¡Ì(4-f(x))/2
x=1+¡Ì(4-f(x), x=1-¡Ì(4-f(x)
(4-f(x))<0
f(x)>4
=f(x)¡Ü 4 (-¡Þ,4] range
domain= all real numbers
deja tienen el sexo en su casa y frotaré su vajina
you just have to translate it and you will see
To find the vertex of the function, you correctly completed the square by rewriting the function in the form y = a(x-h)^2 + k. From there, you correctly identified that the vertex is located at the point (h, k). In this case, h = 1 and k = 4, so the vertex is (1, 4).
To find the axis of symmetry, you can use the x-coordinate of the vertex. The axis of symmetry is a vertical line that passes through the vertex. So, the axis of symmetry for this function is x = 1.
To find the intercepts, you can set y = 0 and solve for x. For the x-intercepts, let y = 0 and solve the equation -x^2 + 2x + 3 = 0. You correctly factored this equation to (x+1)(x-3) = 0, which gives x = -1 and x = 3 as the x-intercepts.
For the y-intercept, you can let x = 0. Plugging this into the function, you correctly found that y = 3. So the y-intercept is (0, 3).
The domain of a function is the set of all possible x-values for which the function is defined. Since this is a polynomial function, the domain is all real numbers.
The range of a function is the set of all possible y-values that the function can take. In this case, since the function is a downward-opening parabola (-x^2 term), the range is all real numbers less than or equal to the y-coordinate of the vertex. So the range is (-∞, 4].
To determine the intervals where the function is increasing or decreasing, calculate the derivative of the function. Taking the derivative of -x^2 + 2x + 3 gives you -2x + 2.
The function is increasing when the derivative is positive, and decreasing when the derivative is negative. Set -2x + 2 > 0 and solve for x to find the intervals where the function is increasing. Similarly, set -2x + 2 < 0 and solve for x to find the intervals where the function is decreasing.
I hope this helps clarify the remaining parts (b)(f), and (g) for you!