Question

What is the vertex of the quadratic function below?

Answer 1

Answer:

Option A: (4, -15).

Step-by-step explanation:

Given the quadratic function, y = x² - 8x + 1, where a = 1, b = -8, and c = 1:

Solve for the x-coordinate of the vertex:

We can use the following equation to solve for the x-coordinate of the vertex:

$\displaystyle\mathsf{x\:=\:\frac{-b}{2a}}$

Substitute the given values into the formula:

$\displaystyle\mathsf{x\:=\:\frac{-b}{2a}\:=\:\frac{-(-8)}{2(1)}\:=\:\frac{8}{2}\:=\:4}$

Hence, the x-coordinate of the vertex is 4.

Solve for the y-coordinate of the vertex:

Next, substitute the x-coordinate of the vertex into the given quadratic function to solve for its corresponding y-coordinate:

y = x² - 8x + 1

y = (4)² - 8(4) + 1

y = 16 - 32 + 1

y = -15

Therefore, the vertex of the given quadratic function, y = x² - 8x + 1, is: x = 4, y = -15, or (4, -15).  Thus, the correct answer is Option A: (4, -15).

Answer 2

Answer:

A. (4, -15)

Step-by-step explanation:

$\underline{\mathrm{Parabola\:equation\:in\:polynomial\:form}}$

$\mathrm{The\:vertex\:of\:an\:up-down\:facing\:parabola\:of\:the\:form}\:y=ax^2+bx+c\:\mathrm{is}\:x_v=-\frac{b}{2a}$

$\mathrm{The\:parabola\:params\:are:}$

$a=1,\:b=-8,\:c=1$

$x_v=-\frac{b}{2a}$

$x_v=-\frac{\left(-8\right)}{2\cdot \:1}$

$x_v=4$

$\mathrm{Plug\:in}\:x_v=4\:\mathrm{to\:find\:the}\:y_v\mathrm{value}$

$y_v=-15$

$\mathrm{Therefore\:the\:parabola\:vertex\:is}$

$\left(4,\:-15\right)$

$\mathrm{If}\:a$

$\mathrm{If}\:a>0,\:\mathrm{then\:the\:vertex\:is\:a\:minimum\:value}$

$a=1$

$\mathrm{Vertex\:of}\:x^2-8x+1:\quad \mathrm{Minimum}\space\left(4,\:-15\right)$