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pochemuha
2 years ago
13

What is the vertex of the quadratic function below?

Mathematics
2 answers:
zysi [14]2 years ago
3 0

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:

<h3><u>Solve for the x-coordinate of the vertex:</u></h3>

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.

<h3><u>Solve for the y-coordinate of the vertex:</u></h3>

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).

hodyreva [135]2 years ago
3 0

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)

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