Identify the Vertex, y-intercept, and x-intercept(s) (Could someone show me step-by-step)

Question

f(x) = x^2 - 14x + 45

Vertex:

y-intercept:

Both x intercepts:

Answer 1

f(x) = x² - 14x + 45

given in the standard form of: Ax² + Bx + C;

where A = 1, B = -14 and C = 45

1. (x,y), the x value of the vertex is x = -2a/b, then use that value to put into f(x) to solve for the y coordinate.

x = (-2)(1)/-14 = -2/-14 = 1/7 (1/7, y)

to find the y coordinate, plug in x = 1/7 ≅.14≅

(1/7)² - 14(1/7) + 45

(1/49) - 2 + 45 ≅ 43.02

So the Vertex is (0.14, 43.02)

1. to find the y intercept, just plug x = 0 into the equation, so 0² - 14(0) + 45 = 45, y intercept = 45
2. to find the x intercepts, that is where the y = 0 (or f(x) = 0), so you could set f(x)=0 and factor the equation, set each part = 0 and solve for x.
3. 0 = (x+9)(x+5)
4. x + 9 = 0; x = -9 and x + 5 = 0; x = -5
5. so the x intercepts are at x = -9 and x = -5

Does this process help you understand how to get the answers?

(Please note there is more than one way to get these answers, the other popular way is to complete the square and put it in vertex format.)

Answer 2

Vertex - The x coordinate of the vertex is at -b/2a = 14/2=7

To find the y coordinate substitute 7 for x y = (7^2) -14(7) +45 = 49-98+45 = -4

So vertex is at (7,-4)

Y intercept To find the y intercept set x equal to zero since x will always be zero at the point the graph crosses

the y intercept so y intercept is 45

X intercepts Solve x^2 -14x +45 =0 (x-9)(x-5)=0 x= 9 or 5

So x intercepts are 9 and 5