I have two questions, please help.
1. Uss completing the square to describe the graph of the following function. Support your answers graphically.
f(x)= -2x^2+4x+7
please show work.
2. Can someone help me to graph this function?
f(x)=x^2-6x+5
Then find the
a. vertex
b. axis of symmetry
c. intercepts, if any
f(x) = -2(x^2-2x) + 7
= -2(x^2-2x+1) + 7 + 2
= -2(x-1)^2 + 9
f(x) = x^2-6x+5
= x^2-6x+9 + 5-9
= (x-3)^2 - 4
now you know all there is to know about the parabola.
Sure, I can help you with both of your questions.
1. Completing the Square to Describe the Graph:
To use completing the square, we want to rewrite the function in the form (x-h)^2 + k, where (h,k) represents the vertex of the parabola.
For the function f(x) = -2x^2 + 4x + 7, we start by factoring out the common factor of -2 from the first two terms:
f(x) = -2(x^2 - 2x) + 7
Now we focus on completing the square within the parentheses. We take half of the coefficient of the x term (-2), square it, and add it inside the parentheses. In this case, (-2/2)^2 = 1:
f(x) = -2(x^2 - 2x + 1) - 2(1) + 7
Next, we simplify inside the parentheses and combine like terms outside the parentheses:
f(x) = -2(x - 1)^2 + 5
Therefore, the graph would be a downward-opening parabola with vertex (1, 5).
To support the answer graphically, you can plot the vertex (1, 5) on a coordinate plane and also plot a few other points to get a better sense of the graph. For example, you can substitute some values of x into the original function and calculate the corresponding y-values, and plot those points as well.
2. Graphing and Finding the Vertex, Axis of Symmetry, and Intercepts:
For the function f(x) = x^2 - 6x + 5:
a. Vertex: To find the vertex, we can use the formula x = -b / (2a), where a and b are the coefficients of x^2 and x respectively. In this case, a = 1 and b = -6. Calculating:
x = -(-6) / (2 * 1) = 6 / 2 = 3
To find the y-value of the vertex, substitute this value of x into the function:
f(3) = (3)^2 - 6(3) + 5 = 9 - 18 + 5 = -4
So, the vertex is (3, -4).
b. Axis of Symmetry: The axis of symmetry is a vertical line that passes through the vertex and divides the parabola into two symmetrical halves. For any parabola in the form f(x) = ax^2 + bx + c, the equation of the axis of symmetry is given by x = -b / (2a). In our case, the axis of symmetry is x = 3.
c. Intercepts: To find the x-intercepts, set f(x) equal to zero and solve for x:
0 = x^2 - 6x + 5
Using the quadratic formula, we get:
x = (-(-6) ± √((-6)^2 - 4(1)(5))) / (2 * 1)
Simplifying, we have:
x = (6 ± √(36 - 20)) / 2
x = (6 ± √16) / 2
x = (6 ± 4) / 2
This gives us two possible x-intercepts:
x = 5 and x = 1
To find the y-intercept, substitute x = 0 into the function:
f(0) = (0)^2 - 6(0) + 5 = 5
So, the y-intercept is (0, 5).
Now, to graph the function, plot the vertex (3, -4), draw the axis of symmetry at x = 3, plot the x-intercepts at (5, 0) and (1, 0), and plot the y-intercept at (0, 5).