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3 years ago

Three points are colinear. always sometimes never

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Three or more points are collinear, if slope of any two pairs of points is same. With three points A, B and C, three pairs of points can be formed, they are: AB, BC and AC. If Slope of AB = slope of BC = slope of AC, then A, B and C are collinear points

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Jan'ai was asked to determine the minimum for a function with zeros located at –1 and 5, which also has a y-intercept of (0, –25

sergey [27]

The first error in Jan'ai's work in determining the considered function is given by: Option D: She incorrectly determined the x-coordinate of the vertex.

<h3>What are the coordinates of vertex for a quadratic function?</h3>

For a quadratic function of the form $y = ax^2 + bx + c$ , its vertex form is obtained as:

$y = ax^2 + bx + c\\y =a(x^2 + bx/a) + c\\y = a(x^2 + 2(b/2a)x + (b/2a)^2 -(b/2a)^2 )+ c\\y = a(x^2 + 2(b/2a)x + (b/2a)^2) -a \times (b/2a)^2 + c\\\\y = a(x+b/2a)^2 - a \times (b/2a)^2 + c$

For the form $y = a(x-h)^2 + k$ , the vertex has coordinates (h, k)

Thus, for the obtained equation $y = a(x+b/2a)^2 - a \times (b/2a)^2 + c$ , we get the coordinates of vertex as:

$h = -b/2a$ , $k = c - a\times(b/2a)^2$

Thus, the coordinates of vertex of $y = ax^2 + bx + c$ is:

$(h,k) = (-b/2a, c - a \times (b/2a)^2 )$

The missing steps of work of Jan'ai are:

Begin to write a function in factored form. $f(x) = a(x+1)(x-5)$
Substitute x = 0, y = f(x) = -25 to determine a. $-25 = a(0+1)(0-5)$
Simplify and solve to find a. $a=5$
Rewrite the function. $f(x) = 5(x+1)(x-5)$
Rewrite in standard form. $f(x) = 5x^2-20x-25$
Find the x-coordinate of the vertex. $x = -20/2(5) = -20/10; x = -2$
Find the y-coordinate of the vertex.

$y = 5x^2-20x-25$

$y = 5(-2)^2-20(-2)-25$

$y = 35$

so (-2,35) is the coordinate of the vertex, which denotes the minimum.

So, as we see, in the 5th step, Jan'ai had the quadratic function $f(x) = 5x^2-20x-25$ ,

Comparing this to $f(x) = ax^2 + bx + c$ , we get a = 5, b = -20, c = -25

The vertex's x-coordinate will be on -b/2a = -(-20)/ 2(5) = 20/10 = 2

But Jan'ai didn't putted that negative sign before b. in the 6th step.

Thus, the first error in Jan'ai's work in determining the considered function is given by: Option D: She incorrectly determined the x-coordinate of the vertex.

Learn more about the vertex form of a quadratic equation here:

brainly.com/question/9912128

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