Quadratic graphs

For any quadratic written as $y=a{x}^2+bx+c$, the number $a$ controls the vertical stretch or compression of the parabola.

What decides the width is the absolute value $|a|$:

• If $|a|$ is larger, the graph is stretched more vertically, so it becomes narrower.
• If $|a|$ is closer to $0$, the graph is vertically squashed, so it becomes wider.

This idea fits into MYP work on Functions and Quadratic Relationships.

Read the coefficient in front of $x^2$ in each option:

• Option A: $y=-8x^2+6x+2⟹a=-8$
• Option B: $y=3x^2-12⟹a=3$
• Option C: $y=-x^2-4x+1⟹a=-1$
• Option D: $y=0.25x^2+8x⟹a=0.25$

To find the narrowest parabola, compare $|a|$ for each function:

$\begin{array}{rl}|a_A|& =|-8|=8\\ |a_B|& =|3|=3\\ |a_C|& =|-1|=1\\ |a_D|& =|0.25|=0.25\end{array}$

The largest absolute value corresponds to the narrowest graph.

Now remove choices with smaller $|a|$ values:

• Option D has the smallest magnitude ($0.25$), so it will be the widest.
• Option C has magnitude $1$, matching the basic width of $y=x^2$.
• Option B has magnitude $3$, which is narrower than the standard parabola, but still not the narrowest here.

Because $8>3>1>0.25$, the equation with $a=-8$ has the largest magnitude. That means the strongest vertical stretch and therefore the narrowest parabola.

To compare widths visually, type each quadratic into a graphing display calculator (GDC) and view them together. A narrower parabola looks steeper: it moves up or down more quickly as you move away from its axis of symmetry.

From the display, $y=-x^2-4x+1$ and $y=0.25x^2+8x$ look much flatter than the others. That matches their small leading-coefficient magnitudes ($|-1|=1$ and $|0.25|=0.25$). So these two can be eliminated.

Now compare $y=-8x^2+6x+2$ with $y=3x^2-12$. On the GDC, the graph of $y=-8x^2+6x+2$ appears more tightly pinched than $y=3x^2-12$. Even though one opens downward and the other opens upward, the larger coefficient magnitude ($8$ compared to $3$) gives the narrower shape.

The visual comparison agrees with the coefficient check: the function with leading coefficient $-8$ is the steepest and therefore gives the narrowest graph.

You are being asked which quadratic form lets you spot the axis of symmetry and the maximum or minimum value (the vertex’s $y$-value) immediately from the equation.

This is part of Functions and Quadratic Relationships. A key skill is knowing which algebraic form shows which graph features without extra work.

A linear form describes a straight line (for example, $y=mx+c$). A line has no vertex and no axis of symmetry like a parabola does, so this choice can be ruled out right away.

Standard form is:

$f(x)=a{x}^2+bx+c$

It’s widely used, but the axis of symmetry is not visible immediately—you must compute $x=-\frac{b}{2a}$. Then you still need to substitute that $x$ back into the function to get the max/min value. Since it takes multiple calculations, it isn’t the most direct choice.

Factorised (intercept) form is:

$f(x)=a(x-p)(x-q)$

This makes the $x$-intercepts $p$ and $q$ easy to see. But the axis of symmetry is found by taking their midpoint: $x=\frac{p+q}{2}$. Because you must do that extra step, it’s not the most efficient form for reading the axis and max/min directly.

Vertex form is built to display the turning point. It is:

$f(x)=a(x-h{)}^2+k$

From this layout, you can read off:

1. The vertex: $(h,k)$
2. The axis of symmetry: $x=h$
3. The max/min value: $y=k$

No extra algebra is required to identify these features.

Since $h$ tells you the symmetry line and $k$ tells you the maximum/minimum value straight away, the vertex form is the most efficient.

We need the quadratic form where the axis of symmetry (the vertex’s $x$-value) and the maximum or minimum value (the vertex’s $y$-value) can be identified just by inspection, not by doing extra steps.

Look at Option C. A linear form represents a straight line (e.g., $y=mx+c$), not a parabola. Since the question is about quadratic graphs with a vertex, this option does not fit.

Now check Option A. Standard form is:

$f(x)=a{x}^2+bx+c$

Although it clearly shows the $y$-intercept ($c$), the symmetry line requires calculating $x=-\frac{b}{2a}$. Because you can’t read the vertex directly, it’s not the most efficient for this task.

Consider Option B. Factorised form is:

$f(x)=a(x-p)(x-q)$

You immediately see the intercepts $p$ and $q$, but the vertex location comes from finding the midpoint between them. Since that still involves computation, it isn’t the most direct option.

That leaves Option D. Vertex form is:

$f(x)=a(x-h{)}^2+k$

Here the vertex is $(h,k)$ and the axis of symmetry is $x=h$. Both features can be read straight from the equation with no calculations.

To answer this, we need the meaning of the $x$-intercepts: they are the points where the graph meets the $x$-axis. Any point on the $x$-axis has $y=0$, so at an $x$-intercept we always have $y=0$.

This fits within Functions and Quadratic Relationships, where choosing the most useful quadratic form is an important skill.

A Linear equation describes a straight line (for example, $y=mx+c$). The question is about a quadratic function, which graphs as a parabola, so a linear form does not match the type of function being discussed.

The Standard form is:

$y=a{x}^2+bx+c$

This form is very familiar, and it makes the $y$-intercept easy to see (it is $c$). But the $x$-intercepts are not immediately visible; you would normally need to factor the quadratic or apply the quadratic formula first, which adds extra work.

The Vertex form is:

$y=a(x-h{)}^2+k$

This is excellent for reading off the vertex (turning point) at $(h,k)$. However, the $x$-intercepts are not shown directly in this form, so you would still need additional algebra to find them.

The Factorised (intercept) form is:

$y=a(x-p)(x-q)$

Here, $p$ and $q$ are the roots. Since $y=0$ when $x=p$ or $x=q$, you can immediately identify the $x$-intercepts as $(p,0)$ and $(q,0)$. That is why this form is usually the most efficient for finding $x$-intercepts.

We are choosing the quadratic form that lets us find the $x$-intercepts of a quadratic function most efficiently. An $x$-intercept happens exactly where the graph crosses the $x$-axis, meaning $y=0$ at those points.

A quadratic has degree 2, so it includes an $x^2$ term. A Linear equation is degree 1 and represents a straight line, not a parabola. So this choice cannot be the best quadratic form for locating intercepts.

Standard form is:

$y=a{x}^2+bx+c$

It’s helpful because the constant term $c$ gives the $y$-intercept quickly. But to get the $x$-intercepts you generally must factor the expression or use the quadratic formula, which means more steps than simply reading them off.

Vertex form is:

$y=a(x-h{)}^2+k$

This form makes the vertex at $(h,k)$ obvious. To find $x$-intercepts, though, you still have to set $y=0$ and solve for $x$ (often using square roots). Since it doesn’t display the intercepts immediately, it isn’t the most efficient here.

Factorised form is:

$y=a(x-p)(x-q)$

By the Null Factor Law, if $y=0$ then $(x-p)=0$ or $(x-q)=0$. That gives $x=p$ and $x=q$ right away. Because the intercepts can be read directly without extra solving, Factorised is the most efficient choice.