Algebra Practice Test 2025 – 400 Free Practice Questions to Pass the Exam

Question: 1 / 400

What is the vertex of the parabola defined by the equation y = x² - 6x + 8?

(2, -2)

(3, -1)

To find the vertex of the parabola defined by the equation \( y = x^2 - 6x + 8 \), we can use the method of completing the square or applying the vertex formula. The vertex form of a parabola is given by \( y = a(x-h)^2 + k \), where the vertex is the point \( (h, k) \).

First, we identify the coefficients in the equation \( y = ax^2 + bx + c \):

- \( a = 1 \) (the coefficient of \( x^2 \))

- \( b = -6 \)

- \( c = 8 \)

The x-coordinate of the vertex can be found using the formula \( h = -\frac{b}{2a} \):

\[

h = -\frac{-6}{2 \cdot 1} = \frac{6}{2} = 3

\]

Next, we substitute \( h \) back into the original equation to find the y-coordinate \( k \):

\[

k = (3)^2 - 6(3) + 8 = 9 - 18 + 8 = -1

\]

Thus, the vertex of the

Get further explanation with Examzify DeepDiveBeta

(0, 8)

(1, 1)

Next Question

Report this question

Download Algebra Practice Test PDF Study Guide
Subscribe

Get the latest from Examzify

You can unsubscribe at any time. Read our privacy policy