Find the vertex form of the function. Then find each of the following.
(A) Intercepts
(B) Vertex
(C) Maximum or minimum and what are the values
(D) Range
s(x)= x^2-8x+7
To find the vertex form of the function, we complete the square.
s(x) = x^2 - 8x + 7
First, we need to factor out the coefficient of the x^2 term, which is 1.
s(x) = (x^2 - 8x) + 7
Next, take half of the coefficient of the x term, which is -8 (-8/2 = -4), and square it (16).
s(x) = (x^2 - 8x + 16) + 7 - 16
Now, rewrite the expression inside parentheses as a binomial squared.
s(x) = (x - 4)^2 + 7 - 16
Simplify the expression within the parentheses.
s(x) = (x - 4)^2 - 9
Therefore, the vertex form of the function is s(x) = (x - 4)^2 - 9.
(A) To find the x-intercepts, we set s(x) = 0 and solve for x.
0 = (x - 4)^2 - 9
(x - 4)^2 = 9
Take the square root of both sides.
x - 4 = ±3
Solve for x.
x = 4 + 3 or x = 4 - 3
x = 7 or x = 1
The x-intercepts are (7, 0) and (1, 0).
(B) The vertex of the function is found by taking the opposite of the x-coordinate of the binomial squared and the y-coordinate is the constant term outside the parentheses.
Vertex = (4, -9)
(C) Since the coefficient of the x^2 term is positive (positive 1 in this case), the parabola opens upwards, which means the vertex is a minimum point. The minimum value of the function is the y-coordinate of the vertex.
Minimum value = -9
(D) The range of the function is all real numbers greater than or equal to the minimum value.
Range = y ≥ -9
To find the vertex form of the function s(x) = x^2 - 8x + 7, we need to complete the square:
s(x) = (x - 4)^2 - 16 + 7
s(x) = (x - 4)^2 - 9
Now let's find each of the following:
(A) Intercepts:
To find the x-intercepts, we set s(x) = 0 and solve for x:
(x - 4)^2 - 9 = 0
(x - 4)^2 = 9
(x - 4) = ±√9
(x - 4) = ±3
x = 4 ± 3
x = 7, 1
So the x-intercepts are (7, 0) and (1, 0).
To find the y-intercept, we substitute x = 0 into the function:
s(0) = 0^2 - 8(0) + 7
s(0) = 7
So the y-intercept is (0, 7).
(B) Vertex:
The vertex form of the function s(x) = (x - h)^2 + k tells us that the vertex is given by (h, k). From the completed square form, we can see that the vertex is (4, -9).
(C) Maximum or minimum and values:
Since the coefficient of the x^2 term is positive (1), the parabola opens upwards, which means it has a minimum value. The minimum value can be found using the y-coordinate of the vertex, which is -9.
(D) Range:
The range of the function s(x) = x^2 - 8x + 7 is all real numbers greater than or equal to the minimum value, -9. So the range is (-∞, -9]
To find the vertex form of a quadratic function, we need to complete the square. Let's start by rewriting the given function:
s(x) = x^2 - 8x + 7
Step 1: Move the constant term to the right side of the equation:
s(x) - 7 = x^2 - 8x
Step 2: Complete the square for the quadratic:
s(x) - 7 + 16 = x^2 - 8x + 16
Step 3: Factor the perfect square trinomial:
s(x) + 9 = (x - 4)^2
Step 4: Rewrite the equation in vertex form:
s(x) = (x - 4)^2 - 9
Now that we have the equation in vertex form, we can find the answers to the given questions.
(A) Intercepts:
To find the x-intercepts, we need to set the function equal to zero and solve for x:
0 = (x - 4)^2 - 9
(x - 4)^2 = 9
Taking the square root of both sides:
x - 4 = ±√9
x - 4 = ±3
x = 4 ± 3
x = 7 or x = 1
So the x-intercepts are x = 7 and x = 1.
To find the y-intercept, we substitute x = 0 into the function:
s(0) = (0 - 4)^2 - 9
s(0) = 16 - 9
s(0) = 7
So the y-intercept is y = 7.
(B) Vertex:
The vertex form of the quadratic function is in the form s(x) = a(x - h)^2 + k, where (h, k) represents the vertex.
Comparing with our equation s(x) = (x - 4)^2 - 9, we can see that the vertex is (h, k) = (4, -9).
(C) Maximum or minimum and its value:
Since the coefficient of the squared term is positive (1 in this case), the parabola opens upwards, meaning it has a minimum. The value of the minimum can be found directly from the vertex coordinates. In this case, the minimum value is -9.
(D) Range:
The range of the quadratic function is the set of all possible y-values. Since the parabola is opening upwards and the minimum value is -9, the range is all real numbers greater than or equal to -9. In interval notation, this can be written as (-∞, -9] or in set notation as {y | y ≥ -9}.