Given that the parent function, y = x², has a leading coefficient of 1, the
following are the comparisons of the equations with the parent functions;
[tex]\begin{array}{|c|cc|}Steeper \, than \, y = x^2&&Less \, Steep \ than \, y = x^2 \\y = 2 \cdot x^2&&y = \frac{1}{2} \cdot x^2 \\y = \left(2 \cdot x \right)^2&&y = \left(\frac{1}{2} \cdot x \right)^2 \end{array}\right][/tex]
The parent function is y = x²
The leading coefficient of the parent function = 1
The steepness of a quadratic equation is given by the value of the
leading coefficient.
When the leading coefficient is larger than 1, we have;
The function is steeper than y = x²
When the leading coefficient is a fraction larger than 0 but lesser than 1, we have;
The function is less steep than y = x²
Which gives;
[tex]\begin{array}{|c|cc|}Steeper \, than \, y = x^2&&Less \, Steep \ than \, y = x^2 \\y = 2 \cdot x^2&&y = \frac{1}{2} \cdot x^2 \\y = \left(2 \cdot x \right)^2&&y = \left(\frac{1}{2} \cdot x \right)^2 \end{array}\right][/tex]
The function y = -x² is inverted has the same steepness as y = x², but will not be used.
Learn more about the properties of quadratic functions here:
https://brainly.com/question/10989920
