Question

the graph of $y = x^2$ is shown on each grid in parts (a) and (b) below.
(a) use transformations to get the graph of $y = 2x^2$.
(b) use transformations to get the graph of $y = (-2x)^2$.

Explanation:

Part (a)

Step1: Recall vertical stretch transformation

For a function \( y = f(x) \), the transformation \( y = a f(x) \) with \( a>1 \) is a vertical stretch by a factor of \( a \). Here, \( f(x)=x^{2} \) and \( a = 2 \), so we apply a vertical stretch to \( y=x^{2} \) by a factor of 2.

Step2: Apply the transformation to key points

• For \( y=x^{2} \), when \( x = 0 \), \( y=0 \); when \( x = 1 \), \( y = 1 \); when \( x=- 1 \), \( y = 1 \); when \( x = 2 \), \( y=4 \); when \( x=-2 \), \( y = 4 \).
• After vertical stretch by factor 2, the new \( y \)-values are: when \( x = 0 \), \( y = 0 \); when \( x = 1 \), \( y=2\times1 = 2 \); when \( x=-1 \), \( y = 2 \); when \( x = 2 \), \( y=2\times4=8 \); when \( x=-2 \), \( y = 8 \). But looking at the grid, we can also think in terms of the shape: the parabola \( y = 2x^{2} \) will be narrower than \( y=x^{2} \) because the coefficient of \( x^{2} \) is greater than 1, and it's a vertical stretch (since the coefficient is outside the function argument). So we take the graph of \( y=x^{2} \) and stretch it vertically by a factor of 2 (each \( y \)-coordinate of a point on \( y=x^{2} \) is multiplied by 2).

Part (b)

Step1: Simplify the function

First, simplify \( y=(-2x)^{2} \). Using the property \( (ab)^{n}=a^{n}b^{n} \), we have \( y=(-2)^{2}x^{2}=4x^{2} \). But also, note that \( (-2x)^{2}=(2x)^{2} \) because squaring removes the negative sign. The transformation for \( y = f(kx) \) with \( |k|>1 \) is a horizontal compression by a factor of \( \frac{1}{|k|} \). Here, \( f(x)=x^{2} \) and \( k = 2 \) (since \( (-2x)^{2}=(2x)^{2} \)), so we have a horizontal compression by a factor of \( \frac{1}{2} \) (or equivalently, a vertical stretch by a factor of 4, but horizontal compression is also a valid way to look at it).

Step2: Apply the transformation to key points

• For \( y=x^{2} \), when \( x = 0 \), \( y = 0 \); when \( x = 1 \), \( y=1 \); when \( x=-1 \), \( y = 1 \); when \( x = 2 \), \( y = 4 \); when \( x=-2 \), \( y = 4 \).
• For \( y = 4x^{2} \) (which is the same as \( y=(-2x)^{2} \)), when \( x = 0 \), \( y = 0 \); when \( x=\frac{1}{2} \), \( y=4\times(\frac{1}{2})^{2}=4\times\frac{1}{4} = 1 \); when \( x=-\frac{1}{2} \), \( y = 1 \); when \( x = 1 \), \( y=4\times1=4 \); when \( x=-1 \), \( y = 4 \). So compared to \( y=x^{2} \), the graph of \( y = 4x^{2} \) (or \( y=(-2x)^{2} \)) is narrower. Alternatively, thinking in terms of \( y=(-2x)^{2} \), since \( (-2x)^{2}=4x^{2} \), it's a vertical stretch by a factor of 4, or a horizontal compression by a factor of \( \frac{1}{2} \) (because \( y=(2x)^{2} \) means we replace \( x \) with \( 2x \), so for a point \( (x,y) \) on \( y=x^{2} \), the corresponding point on \( y=(2x)^{2} \) is \( (\frac{x}{2},y) \), which is a horizontal compression by \( \frac{1}{2} \)). So we take the graph of \( y=x^{2} \) and either compress it horizontally by a factor of \( \frac{1}{2} \) (making it narrower) or stretch it vertically by a factor of 4.

Final Answers (for the transformation descriptions)

(a) The graph of \( y = 2x^{2} \) is obtained by vertically stretching the graph of \( y=x^{2} \) by a factor of 2.
(b) The graph of \( y=(-2x)^{2}=4x^{2} \) is obtained by horizontally compressing the graph of \( y=x^{2} \) by a factor of \( \frac{1}{2} \) (or vertically stretching by a factor of 4).

(If we were to describe the graphing steps:
For (a): Plot the points as calculated (or use the vertical stretch) to draw the narrower parabola \( y = 2x^{2} \).
For (b): Plot the points as calculated (or use the horizontal compression/vertical stretch) to draw the narrower parabola \( y = 4x^{2} \) (which is the same as \( y=(-2x)^{2} \)).)

Answer:

Step1: Simplify the function

First, simplify \( y=(-2x)^{2} \). Using the property \( (ab)^{n}=a^{n}b^{n} \), we have \( y=(-2)^{2}x^{2}=4x^{2} \). But also, note that \( (-2x)^{2}=(2x)^{2} \) because squaring removes the negative sign. The transformation for \( y = f(kx) \) with \( |k|>1 \) is a horizontal compression by a factor of \( \frac{1}{|k|} \). Here, \( f(x)=x^{2} \) and \( k = 2 \) (since \( (-2x)^{2}=(2x)^{2} \)), so we have a horizontal compression by a factor of \( \frac{1}{2} \) (or equivalently, a vertical stretch by a factor of 4, but horizontal compression is also a valid way to look at it).

Step2: Apply the transformation to key points

• For \( y=x^{2} \), when \( x = 0 \), \( y = 0 \); when \( x = 1 \), \( y=1 \); when \( x=-1 \), \( y = 1 \); when \( x = 2 \), \( y = 4 \); when \( x=-2 \), \( y = 4 \).
• For \( y = 4x^{2} \) (which is the same as \( y=(-2x)^{2} \)), when \( x = 0 \), \( y = 0 \); when \( x=\frac{1}{2} \), \( y=4\times(\frac{1}{2})^{2}=4\times\frac{1}{4} = 1 \); when \( x=-\frac{1}{2} \), \( y = 1 \); when \( x = 1 \), \( y=4\times1=4 \); when \( x=-1 \), \( y = 4 \). So compared to \( y=x^{2} \), the graph of \( y = 4x^{2} \) (or \( y=(-2x)^{2} \)) is narrower. Alternatively, thinking in terms of \( y=(-2x)^{2} \), since \( (-2x)^{2}=4x^{2} \), it's a vertical stretch by a factor of 4, or a horizontal compression by a factor of \( \frac{1}{2} \) (because \( y=(2x)^{2} \) means we replace \( x \) with \( 2x \), so for a point \( (x,y) \) on \( y=x^{2} \), the corresponding point on \( y=(2x)^{2} \) is \( (\frac{x}{2},y) \), which is a horizontal compression by \( \frac{1}{2} \)). So we take the graph of \( y=x^{2} \) and either compress it horizontally by a factor of \( \frac{1}{2} \) (making it narrower) or stretch it vertically by a factor of 4.

Final Answers (for the transformation descriptions)

(a) The graph of \( y = 2x^{2} \) is obtained by vertically stretching the graph of \( y=x^{2} \) by a factor of 2.
(b) The graph of \( y=(-2x)^{2}=4x^{2} \) is obtained by horizontally compressing the graph of \( y=x^{2} \) by a factor of \( \frac{1}{2} \) (or vertically stretching by a factor of 4).

(If we were to describe the graphing steps:
For (a): Plot the points as calculated (or use the vertical stretch) to draw the narrower parabola \( y = 2x^{2} \).
For (b): Plot the points as calculated (or use the horizontal compression/vertical stretch) to draw the narrower parabola \( y = 4x^{2} \) (which is the same as \( y=(-2x)^{2} \)).)