How do you graph quadratic functions in vertex form?

The standard form of quadratic function is: f(x) = a(x - h)^2 + k, a is different than 0

The graph of f is a parabola whose vertex it is the point (h, k). If a > 0, the parabola opens upward; if a < 0, the parabola opens downward. Furthermore, if |a| is small, the parabola opens more flatly than if |a| is large. It is a general procedure for graphing parabolas whose equations are in standard form:

Example 1:

Graph the the quadratic function f(x) = -2(x - 3)^2 + 8

Solution:

Standard form: f(x) = a(x - h)^2 + k Given function: f(x) = -2(x - 3) + 8

From the give function we have: a= -2; h= 3; k = 8

Step 1. Determine how the parabola opens.

Note that a = -2. Since a < 0, the parabola is open downward.

Step 2. Find the vertex.

The vertex of parabola is at (h, k). because h = 3 and k = 8, the parabola has its vertex at (3, 8).

Step 3. Find the x-intercepts by solving f(x) = 0. Replace f(x) with 0 at

f(x) = -2(x - 3)^2 + 8 and solve for x

0 = -2(x - 3)^2 + 8

2(x - 3)^2 = 8

(x- 3)^2 = 4

x - 3 = square radical 4

x - 3 = 2 or x -3 = -2

x = 5 or x = 1

The x- intercepts are 1 and 5. Thus the parabola passes through the points (1, 0) and (5, 0), this means that parabola intercepts the x-axis at 1 and 5.

Step 4. Find the y-intercept by computing f(0). Replace x with 0 in f(x) = _2(x - 3)^2 + 8

f(0) = -2(0 - 3)^2 + 8

f(0) = -2(9) + 8

f(0) = -10

The y-intercept is -10. Thus the parabola passes through the point (0, -10), this means that parabola intercepts the y-axis at -10.

Step 5. Graph the parabola. With a vertex at (3, 8), x-intercepts at 1 and 5, and a y-intercept at -10. The axis of symmetry is the vertical line whose equation is x = 3.

Example 2:

Graphing a quadratic function in the form f(x) = ax^2 + bx + c

Graph the quadratic function f(x) = -x^2 - 2x + 1

Solution:

Here a = -1, b = -2, and c = 1

Step 1. Determine how the parabola opens. Since a = 1, a < 0, the parabola opens downward.

Step 2. Find the vertex.

We know that x-coordinate of the vertex is x = -b/2a. Substitute a with -1 and b with -2 into the equation for the x-coordinate:

x = - b/2a

x= -(-2)/(2)(-1)

x = -1, so the x-coordinate of the vertex is -1, and the y-coordinate of the vertex will be f(-1). thus the vertex is at ( -1, f(-1) )

f(x) = -x^2 - 2x +1

f(-1) = -(-1)^2 - 2(-1) + 1

f(-1) = -1 + 2 + 1

f(-1) = 2

So the vertex of the parabola is (-1, 2)

Step 3. Find the x-intercepts by solving f(x) = o

f(x) = -x^2 -2x + 1

0 = -x^2- 2x + 1

We can't solve this equation by factoring, so we use the quadratic formula to solve it.

we get to solution: One solution is x = -2.4 and the other solution is 0.4 (approximately). Thus the x-intercepts are approximately -2.4 and 0.4. The parabola passes through ( -2.4, 0) and (0.4, 0)

Step 4. Find the y-intercept by computing f(0).

f(x) = -x^2 - 2x + 1

f(0) = -(0)^2 - 2(0) + 1

f(0) = 1

The y-intercept is 1. The parabola passes through (0, 1).

Step 5. graph the parabola with vertex at (-1, 2), x-intercepts approximately at -2.4 and 0.4, and y -intercept at 1. The line of symmetry is the vertical line with equation

x= -1.