Graphing Quadratic Functions A quadratic function f is given. (a) Express f in standard form. (b

Graphing Quadratic Functions A quadratic function f is...

Key Concepts

Quadratic Functions

Quadratic functions are polynomial functions of degree two, generally represented by the form ax² + bx + c, where a, b, and c are constants and a ? 0. They model parabolic relationships in various contexts and serve as a foundational concept in algebra, illustrating how changes in variables affect the graph’s shape and position.

Standard Form

The standard form of a quadratic function is expressed as a(x - h)² + k, where (h, k) represents the vertex of the parabola. Converting a quadratic function from its general form to standard form simplifies the process of identifying key features such as the vertex and the axis of symmetry.

Vertex

The vertex of a parabola is the point where the function reaches its minimum or maximum value. In the standard form a(x - h)² + k, the vertex is directly given as (h, k). The vertex is a crucial concept as it indicates the turning point of the quadratic function and aids in understanding its graph.

x- and y-intercepts

x-intercepts are points where the quadratic function crosses the x-axis, found by setting f(x) equal to zero and solving the resulting quadratic equation. The y-intercept is the point where the graph crosses the y-axis, obtained by evaluating the function at x = 0. Both intercepts provide important information regarding the position and shape of the graph.

Graphing Parabolas

Graphing a quadratic function involves plotting its key features, including the vertex, axis of symmetry, and intercepts, and then sketching a smooth curve that represents the parabola. Understanding the direction in which the parabola opens, based on the coefficient a, also plays a significant role in creating an accurate graph.

Domain and Range

The domain of a quadratic function typically consists of all real numbers, since there are no restrictions on the input. The range is determined by the vertex and the direction the parabola opens; if it opens upward, the range starts at the vertex's y-value and extends to infinity, whereas if it opens downward, it extends from negative infinity up to the vertex’s y-value. These concepts define the set of possible output values for the function.

Transcript

00:01 So this question first asks us to express our equation f of x in standard form.

00:04 And the way we do that is we're going to need to complete the square.

00:06 The first step to complete in the square is making sure that our x squared coefficient is 1.

00:10 And in this case, it's not 1.

00:11 Our x squared coefficient is actually 2.

00:13 So the way that we amend this is that we need to divide our equation by 2 or factorise out 2.

00:20 So we'll factorise out 2 here and we get x squared minus 10x inside the bracket.

00:25 And we can leave the plus 57 on the outside for now.

00:28 Okay? and then to continue completing the square, we need to figure out what squares to get x squared minus x.

00:35 So we're going to need an x to get the x squared.

00:38 And then to get the minus 10x, it's always going to be half of what the value is.

00:42 So it's going to be minus 5.

00:43 And if you expand this out here, just to check, you get x squared minus 5x, minus 5x.

00:48 And then you also get a plus 25.

00:50 So to cancel that out, we're going to minus 25.

00:53 And we're keeping our plus 57.

00:55 Since it's times by 2, though, the minus 25, we're going to need to minus 25.

00:59 Again or minus 2 times by 25.

01:02 And that's going to leave us with 2 times by x minus 5, all squared, plus 7.

01:09 Ok.

01:11 Then for part b, it asks us to find the vertex and the x and y intercept.

01:18 So we can get the vertex from our standard form equation.

01:21 Our x value for our vertex is going to be 5, because it's going to be the opposite sign to whatever numbers right here.

01:27 And then for our y value, it's going to be 7.

01:30 Because it's going to be the same sign and it's going to be the value right here.

01:35 So our vertex is 5, 7.

01:39 Okay, so find our x and y intercepts.

01:41 We'll start with our x intercepts first.

01:43 To find our x intercepts, we need to set f of x or y to 0.

01:46 We've already completed the square, so we can use that to solve our equation.

01:51 We've got x minus 5 all squared plus 7 equals 0...

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