Math  /  Algebra

QuestionGraph f(x)=x28x+15f(x)=x^{2}-8 x+15 below by first selecting the correct shape, clicking the vertex, then clicking an xx-Intercept. Clear All Draw:

Studdy Solution

STEP 1

1. The function f(x)=x28x+15 f(x) = x^2 - 8x + 15 is a quadratic function.
2. The graph of a quadratic function is a parabola.
3. The parabola opens upwards since the coefficient of x2 x^2 is positive.
4. The vertex and x x -intercepts can be found using algebraic methods.

STEP 2

1. Identify the shape of the graph.
2. Find the vertex of the parabola.
3. Find the x x -intercepts of the parabola.
4. Plot the vertex and one x x -intercept on the graph.

STEP 3

Identify the shape of the graph. Since the function is quadratic, the graph is a parabola. The coefficient of x2 x^2 is positive, so the parabola opens upwards.

STEP 4

Find the vertex of the parabola. The vertex form of a quadratic function is f(x)=a(xh)2+k f(x) = a(x-h)^2 + k , where (h,k) (h, k) is the vertex. For a standard form ax2+bx+c ax^2 + bx + c , the vertex x x -coordinate is given by:
h=b2a h = -\frac{b}{2a}
For f(x)=x28x+15 f(x) = x^2 - 8x + 15 , a=1 a = 1 , b=8 b = -8 . Calculate h h :
h=82×1=82=4 h = -\frac{-8}{2 \times 1} = \frac{8}{2} = 4

STEP 5

Substitute x=4 x = 4 into the function to find the y y -coordinate of the vertex:
f(4)=428×4+15 f(4) = 4^2 - 8 \times 4 + 15 f(4)=1632+15 f(4) = 16 - 32 + 15 f(4)=1 f(4) = -1
The vertex is (4,1) (4, -1) .

STEP 6

Find the x x -intercepts by setting f(x)=0 f(x) = 0 and solving for x x :
x28x+15=0 x^2 - 8x + 15 = 0
Factor the quadratic equation:
(x3)(x5)=0 (x - 3)(x - 5) = 0

STEP 7

Solve for x x to find the x x -intercepts:
x3=0orx5=0 x - 3 = 0 \quad \text{or} \quad x - 5 = 0 x=3orx=5 x = 3 \quad \text{or} \quad x = 5
The x x -intercepts are (3,0) (3, 0) and (5,0) (5, 0) .

STEP 8

Plot the vertex (4,1) (4, -1) and one of the x x -intercepts, say (3,0) (3, 0) , on the graph. The parabola will open upwards with these points.
The graph of f(x)=x28x+15 f(x) = x^2 - 8x + 15 is a parabola with vertex at (4,1) (4, -1) and x x -intercepts at (3,0) (3, 0) and (5,0) (5, 0) .

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