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:
We can use the following equation to solve for the x-coordinate of the vertex:
[tex]\displaystyle\mathsf{x\:=\:\frac{-b}{2a}}[/tex]
Substitute the given values into the formula:
[tex]\displaystyle\mathsf{x\:=\:\frac{-b}{2a}\:=\:\frac{-(-8)}{2(1)}\:=\:\frac{8}{2}\:=\:4}[/tex]
Hence, the x-coordinate of the vertex is 4.
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:
A. (4, -15)
Step-by-step explanation:
[tex]\underline{\mathrm{Parabola\:equation\:in\:polynomial\:form}}[/tex]
[tex]\mathrm{The\:vertex\:of\:an\:up-down\:facing\:parabola\:of\:the\:form}\:y=ax^2+bx+c\:\mathrm{is}\:x_v=-\frac{b}{2a}[/tex]
[tex]\mathrm{The\:parabola\:params\:are:}[/tex]
[tex]a=1,\:b=-8,\:c=1[/tex]
[tex]x_v=-\frac{b}{2a}[/tex]
[tex]x_v=-\frac{\left(-8\right)}{2\cdot \:1}[/tex]
[tex]x_v=4[/tex]
[tex]\mathrm{Plug\:in}\:x_v=4\:\mathrm{to\:find\:the}\:y_v\mathrm{value}[/tex]
[tex]y_v=-15[/tex]
[tex]\mathrm{Therefore\:the\:parabola\:vertex\:is}[/tex]
[tex]\left(4,\:-15\right)[/tex]
[tex]\mathrm{If}\:a<0,\:\mathrm{then\:the\:vertex\:is\:a\:maximum\:value}[/tex]
[tex]\mathrm{If}\:a>0,\:\mathrm{then\:the\:vertex\:is\:a\:minimum\:value}[/tex]
[tex]a=1[/tex]
[tex]\mathrm{Vertex\:of}\:x^2-8x+1:\quad \mathrm{Minimum}\space\left(4,\:-15\right)[/tex]
