need help asap! thanks!

Question

A. Explain the procedure (but don't create the graph) for graphing the quadratic function in vertex form f(x) = 2(x - 3)² - 1 using pencil and paper. Your explanation should show your understanding of using the perfect squares pattern from the vertex. Explain the entire process.

B. Explain the procedure(but don't create the graph) for graphing the quadratic function in factored form f(x) = (x - 2)(x + 4) using pencil and paper. Your explanation should show your understanding of the intercept graphing method. Explain the entire process.

C. Explain how a function graph can be used to help verify that a function is factored correctly.

Answer 1

For graphs in general, it is very helpful to find the vertex point right off the bat, then you can elaborate upon that vertex point for the rest of the graph.

a) Intuitively, for the vertex form f(x) = A(x-H)²+ K, A would be the magnitude of the function, H would be the steps away from the origin in the x-axis (from LEFT to RIGHT, or negative left side of origin and positive right side of origin), and K would then be the steps from the origin in the y-axis(from BOTTOM to TOP, or negative bellow origin positive above origin). For simple solution, you can get the vertex point from the function as H = 3 and K = -1 which is (3,-1) for the vertex point, then you would graph a f(x)=2x² origin at (3,-1). For a longer path, If you assume f(x)=2(x-3)²-1 as f(x) = (x-3)² you would know what's the vertex's x-value would be: (x-3) would solve as x=3. Then for vertex's y-value plug in the x-value found from previous step into the function: f(3)= 2(3-3)²-1 = 0-1 = -1. Lastly, you graph a f(x)=2x² at (3,-1).

b)Recall that a parabola function is symmetrical, you can find the x-value of the vertex by solving the roots: (x-2) --> x=2 (x+4) --> x=-4 and the vertex x-value must be right between these two values, you can find it by calculating (x₁+x₂)/2 ---> (2-4)/2 = -1. Afterwards, you can simply solve for the y-value for the vertex by plugging in x=-1, which is f(-1)= (-3)(3) = -9, that is, the vertex point is at (-1, -9) and you can graph a f(x)=x² at (-1,-9) as the origin.(p.s. you know it is a (1)x² because (ax+h)(bx+k) that a*b = 1)

c) The value of x-intercepts should match with the factored function's number. Take b) for example, (2,0) and (-4,0) would be the x-intercepts for the graph if the function is graphed out. Furthermore, if the graph shows the functions intersects the x-axis once, that means it is a double root, which means your factored function should look somthing like f(x)= (x+K)², which (K, 0) is the only x-intercept you have.

Answer 2

f(x) = 2(x-3)^2 + 1 has vertex (3,1)

f(x) = 2(x-h)^2 + k has vertex (h,k)

for 2(x-3)^2 + 1, h=3 and k=1, so vertex (h,k) is (3,1)

the polynomial is in vertex form, with the x terms completing a square

since 2>0, the parabola opens upward and the vertex is the minimum point of the parabola

the vertex is above the x axis, so there are no x intercepts. zeros or roots are imaginary.

set f(x) = 0, solve for x and there are no real solutions

expand the completed square, 2(x-3)^2 + 1 to get 2x^2 - 12x+ 18 + 1 and the y intercept is the constant term: 18+1 = 19 or the point (0,19). axis of symmetry is x=3, so the point (6,19) is also a point on the parabola. That gives 3 points on the parabola, (3,1), (0,19) and (6,19). Draw a smooth parabolic curve through those points for the graph.

B f(x)= (x-2)(x+4) has x intercepts x=2 and x=-4 or the points (2,0) and (-4,0) the axis of symmetry is x=-1, which is the average of the x intercepts (2-4)/2 =-2/2 = -1

the vertex will have an x coordinate =-1 and be below the x axis with a negative y coordinate. Expand the two factors to find the vertex. f(x) = x^2+2x -8, complete the square: x^2+2x +1 -8-1 = (x+1)^2 -9

the vertex is (-1,-9). That gives 3 points on the parabola, the x intercepts and the vertex (2,0), (-4,0) and (-1,-9) Draw a smooth U shaped curve through the 3 points. the vertex is the minimum point of the parabola

when you complete the square, the vertex should have an x coordinate that is the average of the x intercepts