Find a quadratic function from two points and solve a one-intersection condition

Expert intro: This question has two parts, but both rely on the same idea: plug in known points, form equations, and then use the condition about intersections to build a second equation.

Detailed walkthrough

(a) Find $a$ and $b$

We are given

$y=ax^2+bx+13$

and the graph passes through $(-3,23)$ and $(4,5)$.

Substitute $(-3,23)$:

$23=a(-3)^2+b(-3)+13$

$23=9a-3b+13$

$9a-3b=10 \quad (1)$

Substitute $(4,5)$:

$5=a(4)^2+b(4)+13$

$5=16a+4b+13$

$16a+4b=-8 \quad (2)$

Now solve the system.

From (1):

$3a-b=\frac{10}{3}$

It may be cleaner to eliminate directly. Multiply (1) by 4:

$36a-12b=40$

Multiply (2) by 3:

$48a+12b=-24$

Add:

$84a=16$

$a=\frac{4}{21}$

Substitute into (1):

$9\left(\frac{4}{21}\right)-3b=10$

$\frac{12}{7}-3b=10$

$-3b=\frac{58}{7}$

$b=-\frac{58}{21}$

So

$\boxed{a=\frac{4}{21},\quad b=-\frac{58}{21}}$

(b) Exactly one intersection

The functions are

$f(x)=3x^2+bx+3$

$g(x)=1-6x$

Set them equal:

$3x^2+bx+3=1-6x$

$3x^2+(b+6)x+2=0$

For exactly one point of intersection, this quadratic must have one repeated root, so the discriminant must be zero:

$\Delta=(b+6)^2-4(3)(2)=0$

$(b+6)^2-24=0$

$(b+6)^2=24$

$b+6=\pm 2\sqrt{6}$

$\boxed{b=-6\pm 2\sqrt{6}}$

💡 Pitfall guide

In part (a), the most common issue is arithmetic when substituting the points. Watch the constants carefully: $23-13=10$ and $5-13=-8$. In part (b), remember that “exactly one point of intersection” means the resulting quadratic has a repeated root, so the discriminant must be zero, not positive.

🔄 Real-world variant

If part (b) asked for at least one intersection instead, you would only need the discriminant to be $\ge 0$. If the line were changed to $g(x)=k-6x$, then the value of $b$ for tangency would change because the constant term in the intersection equation would change as well.

🔍 Related terms

quadratic function, discriminant, point of intersection