Q. the vertex of the graph of y = (x – 1)^2 – 5 is

Q: What is the vertex of y = (x - 1)^2 - 5?

The vertex is (1,-5). This is already in vertex form y=(x-h)^2+k with h=1 and k=-5.

Q: What is the axis of symmetry?

The axis of symmetry is the vertical line x=1, since it passes through the vertex (1,-5).

Q: Does the parabola open up or down?

It opens upward because the coefficient of (x-1)^2 is positive (1). positive a in y=a(x-h)^2+k means up.

Q: How do I graph this quickly?

Plot the vertex (1,-5), draw the axis x=1, mark symmetric points (e.g., x=0 gives y=-4, x=2 gives y=-4), then sketch the upward U-shaped curve.

Q: How do I find the vertex from standard form y=ax^2+bx+c?

Use h=-\frac{b}{2a}, then compute k=f(h). Alternatively complete the square to rewrite in vertex form y=a(x-h)^2+k.

Q: How do h and k in y=(x-h)^2+k affect the graph?

h shifts the parabola right by h (left if h<0); k shifts it up by k (down if k<0). Together they move the vertex to (h,k).

Q: What are the x- and y-intercepts?

Y-intercept: at x=0, y=(0-1)^2-5=-4, so (0,-4). X-intercepts solve (x-1)^2-5=0 giving x=1\pm\sqrt{5}, so \left(1\pm\sqrt{5},0\right).

Q: Why is vertex form useful?

Vertex form y=a(x-h)^2+k immediately gives the vertex (h,k), axis of symmetry x=h, direction/stretch from a, and simplifies graphing and translations.

Answer

The parabola is already in vertex form \(y=(x-1)^2-5\), so the vertex is \((1,-5)\).

Detailed Explanation

Problem

Find the vertex of the graph of the quadratic function \( y = (x – 1)^2 – 5 \).

Step-by-step solution

Recognize the vertex form.

A quadratic written as \( y = a(x – h)^2 + k \) is in vertex form. The vertex of the parabola is the point \( (h, k) \).
Compare the given function to vertex form.

The given function is already in vertex form:

\( y = (x – 1)^2 – 5 \).

By comparison, \( a = 1 \), \( h = 1 \), and \( k = -5 \).
Determine the vertex coordinates.

The vertex is \( (h, k) \). Substituting the identified values gives the vertex

\( (1, -5) \).
Optional explanation why this is the vertex.

Because \( (x – 1)^2 \ge 0 \) for all real x, the smallest value of \( y \) occurs when \( (x – 1)^2 = 0 \), i.e., at \( x = 1 \). At that x,

\( y = 0 – 5 = -5 \).

Thus the minimum point (vertex) is \( (1, -5) \). Also, since \( a = 1 > 0 \), the parabola opens upward and the vertex is a minimum.

Answer

The vertex is \( (1, -5) \).

See full solution

Graph

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FAQs

What is the vertex of \(y = (x - 1)^2 - 5\)?

The vertex is \((1,-5)\). This is already in vertex form \(y=(x-h)^2+k\) with \(h=1\) and \(k=-5\).

What is the axis of symmetry?

The axis of symmetry is the vertical line \(x=1\), since it passes through the vertex \((1,-5)\).

Does the parabola open up or down?

It opens upward because the coefficient of \((x-1)^2\) is positive (1). positive \(a\) in \(y=a(x-h)^2+k\) means up.

How do I graph this quickly?

Plot the vertex \((1,-5)\), draw the axis \(x=1\), mark symmetric points (e.g., \(x=0\) gives \(y=-4\), \(x=2\) gives \(y=-4\)), then sketch the upward U-shaped curve.

How do I find the vertex from standard form \(y=ax^2+bx+c\)?

Use \(h=-\frac{b}{2a}\), then compute \(k=f(h)\). Alternatively complete the square to rewrite in vertex form \(y=a(x-h)^2+k\).

How do \(h\) and \(k\) in \(y=(x-h)^2+k\) affect the graph?

\(h\) shifts the parabola right by \(h\) (left if \(h<0\)); \(k\) shifts it up by \(k\) (down if \(k<0\)). Together they move the vertex to \((h,k)\).

What are the x- and y-intercepts?

Y-intercept: at \(x=0\), \(y=(0-1)^2-5=-4\), so \((0,-4)\). X-intercepts solve \((x-1)^2-5=0\) giving \(x=1\pm\sqrt{5}\), so \(\left(1\pm\sqrt{5},0\right)\).

Why is vertex form useful?

Vertex form \(y=a(x-h)^2+k\) immediately gives the vertex \((h,k)\), axis of symmetry \(x=h\), direction/stretch from \(a\), and simplifies graphing and translations.

Find the vertex of the parabola here.
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Q. the vertex of the graph of y = (x – 1)^2 – 5 is

Answer

Detailed Explanation

Problem

Step-by-step solution

Answer

Graph

FAQs

What is the vertex of \(y = (x - 1)^2 - 5\)?

What is the axis of symmetry?

Does the parabola open up or down?

How do I graph this quickly?

How do I find the vertex from standard form \(y=ax^2+bx+c\)?

How do \(h\) and \(k\) in \(y=(x-h)^2+k\) affect the graph?

What are the x- and y-intercepts?

Why is vertex form useful?

Similar Questions

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