Jeffrey is asked to graph f(x)=x^2 - 6x + 5. Below is his graph. Is his graph correct or incorrect?
Prove your answer by discussing the key characteristics of a quadratic function. Include the shape of the graph based on its leading coefficient, calculation of its vertex, maximum or minimum value as well as its x-intercepts. Show your work to receive full credit.

His graph is incorrect, because, Jeffrey drew it with vertex (4 , -4) and x-intercepts (2 , 0) and (5 , 0), but the vertex is (3 , -4) and x-intercepts are (1 , 0) , (5 , 0)

Step-by-step explanation:

The quadratic function y = ax² + bx + c is represented graphically by a parabola

∵ The quadratic is f(x) = x² - 6x + 5

∴ a = 1 , b = -6 and c = 5

∵ The leading coefficient a is positive

∴ The graph of f(x) is a parabola oped upward

∴ The parabola has a minimum vertex

The vertx point of the parabola is (h , k)

∵ h =

∴ h =  

∴ h = 3

- To find k substitute x by 3 in f(x)

∵ f(3) = (3)² - 6(3) + 5 = 9 - 18 + 5

∴ f(3) = -4

∵ k = f(3)

∴ k = -4

That means the coordinates of the vertex of the parabola are (3 , -4)

The x-intercepts means values of x at f(x) = 0

∵ x² - 6x + 5 = 0

- Factorize it into two binomial factors

∵ x² = (x)(x)

∵ 5 = (-1)(-5)

∵ (x)(-1) + (x)(-5) = -x + -5x = -6x

- That means the factors are (x - 1) and (x - 5)

∴ The factors of x² - 6x + 5 are (x - 1) and (x - 5)

∴ (x - 1)(x - 5) = 0

Equate each factor by 0 to find the value of x

∵ x - 1 = 0

- add 1 to both sides

∴ x = 1

∵ x - 5 = 0

- add 5 to both sides

∴ x = 5

∴ The x-intercepts are (1 , 0) and (5 , 0)

That means the parabola intersects the x-axis ate points (1 , 0) and (5 , 0)

Jeffrey drew the graph with vertex (4 , -4) and x-intercepts (2 , 0) and (5 , 0), then his graph is incorrect

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