Which is the graph of f(x)=(x-1)(x+4)?

Asked by maham237 @ 22/12/2020 in Mathematics viewed by 344 People

Answered by maham237 @ 22/12/2020

The parabola opens upward
The x-intercepts at (-4, 0) and (1, 0)
The y-intercept is the point (0, -4)
The axis of symmetry is x = - ³/₂
The minimum value is -6¹/₄
The vertex is (- ³/₂, - 6¹/₄)

Further explanation

The quadratic function is described by the standard equation $\boxed{ \ f(x) = ax^2 + bx + c \ }.$

We have the quadratic function $\boxed{ \ f(x) = (x - 1)(x + 4) \ }.$

Let us configure it to obtain a standard equation.

$\boxed{f(x) = (x - 1)(x + 4)}$

$\boxed{f(x) = x^2 + 4x - x - 4}$

Hence, we get $\boxed{\boxed{ \ f(x) = x^2 + 3x - 4 \ }}.$

We identify the coefficients a, b, and c. For this equation, $\boxed{ \ a = 1, b = 3, and \ c = -4 \ }$
The parabola opens upward because a > 0, resulting in a vertex that is a minimum.
The y-intercept of the quadratic function f(x) = x² + 3x - 4 is (0, c), i.e., the point $\boxed{ \ (0, -4) \ }.$
From $\boxed{ \ f (x) = (x - 1)(x + 4) \ }$ we get the x-intercepts at $\boxed{ \ (-4, 0) \ and \ (1, 0) \ }$
The axis of symmetry is $\boxed{ \ x = h = -\frac{b}{2a} \ }$ , i.e., $\boxed{ \ x = h = -\frac{3}{2(1)} \rightarrow h = -\frac{3}{2} \ }$
The minimum value is $\boxed{ \ k = -\frac{25}{4} = -6\frac{1}{4} \ }$
The vertex is $\boxed{ \ (h, k) \ },$ where $\boxed{ \ k = f(h) \ }$ or $\boxed{ \ k = \frac{b^2 - 4ac}{-4a} \ }$

Finding the minimum value is as follows:

$\boxed{ \ k = f (- \frac{3}{2}) = (- \frac{3}{2})^2 + 3(- \frac{3}{2}) - 4 = -\frac{25}{4} = -6\frac{1}{4} \ }, or$
$\boxed{ \ k = \frac{3^2 - 4(1)(-4)}{-4(1)} = -\frac{25}{4} = -6\frac{1}{4} \ }$

Notes:

The graph of a quadratic function is called a parabola.
When a > 0, the parabola opens upward, resulting in a vertex that is a minimum.
When a < 0, the parabola opens downward, resulting in a vertex that is a maximum.
The value c is the y-intercept of the graph, because a y-intercept is a point on the graph where x is zero. In other words, the graph passes through the point $\boxed{ \ (0, c) \ }.$
From $\boxed{ \ f (x) = (x - x_1)(x - x_2) \ }$ we get x-intercepts at $\boxed{ \ (x_1, 0) \ and \ (x_2, 0) \ }.$
An x-intercept represents a point on the graph where y is zero.
The axis of symmetry represent the line that passes through the vertex of parabola with equation $\boxed{ \ x = h = -\frac{b}{2a} \ }$
The vertex is $\boxed{ \ (h, k) \ },$ where $\boxed{ \ k = f(h) \ }$ or $\boxed{ \ k = \frac{b^2 - 4ac}{-4a} \ }$

Learn more

A line that is not parallel to either the x-axis or the y-axis brainly.com/question/4691222
Finding the y-intercept of the quadratic function f(x) = (x – 6)(x – 2) brainly.com/question/1332667
The midpoint brainly.com/question/3269852

Keywords: which is the graph of f(x) = (x - 1)(x + 4), the x-intercept, quadratic function, a standard equation, the y-intercept, the axis of symmetry, the vertex, parabola, upward, downward

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