Advanced Math · High-Frequency Tested Skill (~35% of Math)
Quadratic Equations & Vertex Form
Key SAT Rule at a Glance
In quadratic modeling on the Digital SAT: standard form ($y = ax^2 + bx + c$) gives the y-intercept ($c$); factored form ($y = a(x - r_1)(x - r_2)$) gives the x-intercepts or roots ($r_1, r_2$); and vertex form ($y = a(x - h)^2 + k$) gives the maximum or minimum turning point ($h, k$) without any algebraic solving.
1. The Three Costumes of a Parabola
Quadratics anchor the Advanced Math domain on the Digital SAT. The same quadratic function can be expressed in three distinct forms, each revealing a different key feature:
Standard Form ($y = ax^2 + bx + c$)
Reveals the $y$-intercept when $x = 0$ ($c$) and direction ($a > 0$ opens up; $a < 0$ opens down).
Factored Form ($y = a(x - r_1)(x - r_2)$)
Reveals the $x$-intercepts (roots or zeros) where $y = 0$. The axis of symmetry lies halfway between them.
Vertex Form ($y = a(x - h)^2 + k$)
Reveals the maximum ($a < 0$) or minimum ($a > 0$) value $k$, occurring at $x = h$.
The Discriminant ($b^2 - 4ac$)
Counts real solutions without solving: $>0$ means two roots, $=0$ means one root, $<0$ means no real roots.
The zero-product property only works when the equation equals zero. If an equation reads $x(x - 4) = 12$, you cannot conclude $x = 12$ or $x - 4 = 12$! You must expand first ($x^2 - 4x - 12 = 0$), then factor to $(x - 6)(x + 2) = 0$, giving $x = 6$ or $x = -2$.
2. Finding Roots and Maxima in Desmos
Whenever a question asks for the roots, vertex, or intersection with a line, Desmos solves it instantly:
- Type the quadratic equation into Desmos line 1: $y = -5(x - 3)^2 + 80$
- Click on the parabola: gray dots appear at the vertex $(3, 80)$ and both $x$-intercepts $(-1, 0)$ and $(7, 0)$.
- Click any gray dot to read coordinates directly.
3. Worked SAT Practice Problem
The height in meters of a projectile launched from a platform is modeled by $h(t) = -5(t - 3)^2 + 80$, where $t$ is the time in seconds after launch. What is the maximum height, in meters, reached by the projectile, and at what time does it reach this height?
View Step-by-Step Explanation
Correct Answer: B (Maximum height 80 meters at $t = 3$ seconds)
The equation is given in vertex form $h(t) = a(t - h)^2 + k$, where the vertex coordinates $(h, k)$ represent the peak of the trajectory. Because $a = -5$ is negative, the parabola opens downward and reaches its maximum at $(3, 80)$. The time of maximum height is $t = 3$ seconds, and the maximum height is $80$ meters.