17.7 Quadratic Curve When It Opens Downwards

🟦 Concept

A quadratic equation can be written in the general form:$$y=ax^2+bx+c$$

The value of aa tells us whether the quadratic curve opens upwards or downwards.

When:$$a<0$$

the curve opens downwards and has an upside-down U-shape ∩.

A downward-opening quadratic has a maximum turning point, which is the highest point on the curve.

Remember:
Negative a → Opens downwards ∩ → Maximum

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🟩 Example Math

Consider the quadratic equation:$$y=-x^2+4x+1$$

Compare it with:$$y=ax^2+bx+c$$

Therefore:$$a=-1,\quad b=4,\quad c=1$$

Since:$$a=-1<0$$

the quadratic curve opens downwards.

Now let’s calculate its important features before plotting the graph.

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1. Find the Turning Point

The x-coordinate of the turning point is:$$x=-\frac{b}{2a}$$

Substitute $a=-1$ and $b=4$:$$x=-\frac{4}{2(-1)}$$$$x=2$$

Now substitute $x=2$ into the original equation:$$y=-(2)^2+4(2)+1$$$$y=-4+8+1$$$$y=5$$

Therefore, the turning point is:$$(2,5)$$

Because the curve opens downwards, this is a maximum turning point.

The axis of symmetry is:$$x=2$$

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2. Find the Roots

The roots are the points where the curve crosses the x-axis.

At the x-axis:$$y=0$$

Therefore:$$-x^2+4x+1=0$$

Multiply by $-1$:$$x^2-4x-1=0$$

This does not factorise easily, so we can use the quadratic formula:$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

For:$$x^2-4x-1=0$$

we have:$$a=1,\quad b=-4,\quad c=-1$$

Therefore:$$x=\frac{4\pm\sqrt{16+4}}{2}$$$$x=\frac{4\pm\sqrt{20}}{2}$$$$x=2\pm\sqrt5$$

So the roots are:$$(2-\sqrt5,0)$$

and$$(2+\sqrt5,0)$$

Approximately:$$(-0.24,0)\quad\text{and}\quad(4.24,0)$$

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3. Find the y-intercept

The y-intercept occurs where the curve crosses the y-axis.

At the y-axis:$$x=0$$

Substitute $x=0$:$$y=-(0)^2+4(0)+1$$$$y=1$$

So the y-intercept is:$$(0,1)$$

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🟨 Now Plot the Quadratic Curve

We now have the key information needed to draw the graph:

Feature	Result
Equation	$y=-x^2+4x+1$
Value of $a$	$-1$ — negative
Direction	Opens downwards ∩
Turning point	$(2,5)$
Type	Maximum
Axis of symmetry	$x=2$
Roots	$(2-\sqrt5,0)$, $(2+\sqrt5,0)$
Approximate roots	$(-0.24,0)$, $(4.24,0)$
y-intercept	$(0,1)$

Plot the important points and join them with a smooth downward-opening curve.

The graph is symmetrical about:$$x=2$$

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🟪 Range

The highest point of the graph is the turning point:$$(2,5)$$

Therefore, the greatest possible value of $y$ is:$$5$$

The rest of the curve extends downwards.

So the range is:$$y\leq5$$

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💡 Exam Tip

Always check the coefficient of $x^2$ first.

For:$$y=ax^2+bx+c$$

a>0a>0 → Opens upwards ∪ → Minimum

a<0a<0 → Opens downwards ∩ → Maximum

This tells you the basic shape of the quadratic before you even plot it.

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⚠️ Common Mistake

Don’t confuse the turning point with the maximum value.

For this example:

Turning point: $(2,5)$

Maximum value: $5$

The turning point is a coordinate, while the maximum value is the highest y-value.

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⭐ Quick Summary

For:$$y=-x^2+4x+1$$

a=−1<0a=-1<0 → curve opens downwards ∩

Turning point: $(2,5)$

Maximum value: $5$

Axis of symmetry: $x=2$

Roots: $x=2\pm\sqrt5$

y-intercept: $(0,1)$

Range:$$y\leq5$$

Remember

Negative aa → Downwards ∩ → Maximum