Q. Which graph represents the function \(f(x) = (x – 5)^2 + 3\)?

Answer

Shifted parabola. From the parent function \(y=x^2\) we have a horizontal shift right 5 and an upward shift 3, so the graph is a parabola opening upward with

– Vertex at \((5,3)\) (minimum point),
– Axis of symmetry \(x=5\),
– No real x-intercepts (since \((x-5)^2+3=0\) gives \((x-5)^2=-3\)),
– y-intercept \(f(0)=(0-5)^2+3=28\).

Thus pick the graph whose parabola has vertex \((5,3)\), opens up, and passes through \((0,28)\).

Detailed Explanation

So, the solution is as follows:

Identify the form and transformations.The function is
\[
f(x)=(x-5)^2+3,
\]
which is the parent parabola \(y=x^2\) shifted right by 5 units and up by 3 units.
Find the vertex.In the form \((x-h)^2+k\), the vertex is \((h,k)\). Here \(h=5\) and \(k=3\), so the vertex is
\[
(5,3).
\]
Axis of symmetry and direction.The axis of symmetry is the vertical line
\[
x=5,
\]
and because the coefficient of \((x-5)^2\) is \(+1\), the parabola opens upward (same width as \(y=x^2\)).
Intercepts and sample points.Compute the intercepts and a few points:
- y-intercept: \(f(0)=(0-5)^2+3=25+3=28\), so \((0,28)\).
- x-intercepts: solve \((x-5)^2+3=0\Rightarrow (x-5)^2=-3\). No real solutions, so there are no x-intercepts.
- Nearby symmetric points: \(f(6)=(6-5)^2+3=1+3=4\) and \(f(4)=4\), so points \((6,4)\) and \((4,4)\). Also \(f(7)=7\) and \(f(3)=7\).
Conclusion — which graph to choose.Choose the graph that is an upward-opening parabola with vertex at \((5,3)\), axis \(x=5\), no x-intercepts, and passing through points such as \((6,4)\), \((4,4)\), and \((0,28)\). That graph represents \(f(x)=(x-5)^2+3\).

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FAQ

What exactly is f(x) = ()–x?

It looks like a typo or missing expression. Ask for the missing part. If it meant f(x) = -x, the function is y = -x (line through origin with slope -1).

If the function is f(x) = -x, what does its graph look like?

straight line through (0,0) with slope -1: for each 1 right, go 1 down. Points: (1,-1), (2,-2), (-1,1).

How do I identify slope and intercept from f(x)=mx+b?

m is slope (steepness, rise/run). b is y-intercept (where line crosses y-axis). For y = -x, m = -1 and b = 0.

How do I find x- and y-intercepts?

y-intercept: set x=0 → y=b. X-intercept: set y=0 → solve 0 = mx + b → x = -b/m (if m ≠ 0). For y=-x, both intercepts are 0.

How can I quickly sketch a linear function?

Plot the y-intercept, then use the slope as rise/run to get another point, draw the line through them. Two points determine the line.

How do I check which of several graphs matches the function?

What is the domain and range of y = -x?

Domain: all real numbers. Range: all real numbers. non-vertical linear function covers every real y.

Is y = -x symmetric?

It is symmetric about the origin (odd function). Rotating the graph 180° about the origin yields the same line.

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