Q. The parabola \(y=3(x-5)^2\) has ____ x-intercept(s).

Q: How many x-intercepts does the parabola y=3(x-5)^2 have?

One. Setting y=0 gives (x-5)^2=0, so x=5 is the only solution (a repeated root), so the graph touches the x-axis at a single point.

Q: Why is the x-intercept a "repeated root" or "double root"?

Because the factor (x-5) appears squared: (x-5)^2=0 has root x=5 with multiplicity 2, meaning the parabola is tangent to the x-axis there rather than crossing it.

Q: What is the vertex and axis of symmetry of y=3(x-5)^2?

The vertex is at (5,0). The axis of symmetry is the vertical line x=5.

Q: Does the parabola cross the x-axis anywhere else?

No. Because the quadratic has a single real solution x=5 (multiplicity 2), the parabola only touches the x-axis at that point and does not cross it.

Q: What is the y-intercept of y=3(x-5)^2?

What is the y-intercept of y=3(x-5)^2?

Q: What is the range and domain of y=3(x-5)^2?

Domain: all real numbers. Range: y\ge 0, since the minimum value at the vertex is 0 and the parabola opens upward.

Q: How do you expand y=3(x-5)^2 to standard polynomial form?

Expand: y=3(x^2-10x+25)=3x^2-30x+75. The quadratic in standard form is y=3x^2-30x+75.

Answer

Set \(y=0\): \(3(x-5)^2=0\). Then \((x-5)^2=0\), so \(x=5\). Thus there is 1 x-intercept at \((5,0)\).

Detailed Explanation

To find x-intercepts set (y) to 0: \[0 = 3(x – 5)^2\]
Divide both sides by 3: \[0 = (x – 5)^2\]
Solve the equation (a perfect square equals zero): \[x – 5 = 0, \\text{ so } x = 5\]
Thus the parabola meets the x-axis at a single point (a repeated root), namely ((5, 0)).

Answer: The parabola has 1 x-intercept (at ((5, 0))).

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FAQs

How many x-intercepts does the parabola \(y=3(x-5)^2\) have?

One. Setting \(y=0\) gives \((x-5)^2=0\), so \(x=5\) is the only solution (a repeated root), so the graph touches the x-axis at a single point.

Why is the x-intercept a "repeated root" or "double root"?

Because the factor \((x-5)\) appears squared: \((x-5)^2=0\) has root \(x=5\) with multiplicity 2, meaning the parabola is tangent to the x-axis there rather than crossing it.

What is the vertex and axis of symmetry of \(y=3(x-5)^2\)?

The vertex is at \((5,0)\). The axis of symmetry is the vertical line \(x=5\).

Does the parabola cross the x-axis anywhere else?

No. Because the quadratic has a single real solution \(x=5\) (multiplicity 2), the parabola only touches the x-axis at that point and does not cross it.

Is the parabola opening up or down, and how does the 3 affect its shape?

It opens upward because the leading coefficient 3 is positive. The factor 3 makes it narrower (steeper) than \(y=(x-5)^2\), scaling vertical distances by 3.

What is the y-intercept of \(y=3(x-5)^2\)?

What is the range and domain of \(y=3(x-5)^2\)?

Domain: all real numbers. Range: \(y\ge 0\), since the minimum value at the vertex is 0 and the parabola opens upward.

How do you expand \(y=3(x-5)^2\) to standard polynomial form?

Expand: \(y=3(x^2-10x+25)=3x^2-30x+75\). The quadratic in standard form is \(y=3x^2-30x+75\).

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