That's correct, with some provisions: the downward curve is not exponential; it's only quadratic; the trend will be upwards for values of $x$ that precede its vertex; and we must take care to distinguish the true curve from the fitted one.
Consider these data:
In both views, there are $30$ $(x,y)$ data values plotted as points. At the left is the quadratic least squares fit in blue. At the right the same fit is shown with the graph of the true underlying model as a dotted line: it is quadratic with a vertex at $(2, 25)$.
As always, interpret a model of the form
$$E[Y] = f(x;\theta)$$
by considering what a unit change in $x$ does to the expectation of $Y$:
$$\Delta_x f(x;\theta) = f(x+1;\theta) - f(x;\theta)\tag{*}.$$
In this case a quadratic model posits that
$$f(x;(\theta_1,\theta_2,\theta_3)) = \theta_1 + \theta_2 x + \theta_3 x^2,$$
from which we compute $(*)$ as
$$( \theta_1 + \theta_2 (x+1) + \theta_3 (x+1)^2 ) - ( \theta_1 + \theta_2 x + \theta_3 x^2) = \theta_2 + \theta_3(2x+1).\tag{**}$$
In the example, the fitted coefficients are $\hat\theta_2 = -0.77$ and $\hat\theta_3=-0.191$: both are negative. Nevertheless, for this range of $x$ from $-10$ to $-3$, the expression $(**)$ is positive. (Evidently it would be negative for any positive $x$.) For the true model, $\theta_2=0.8$ and $\theta_3=-0.2$: it predicts a positive change in $y$ beginning at ay $x \lt 3/2$.
To determine whether this rate of change $(*)$ increases or decreases with $x$, find its first difference, $\Delta_x\left(\Delta_x f(x;\theta)\right)$. In the example, using the simple expression in $(**)$, this works out to
$$(\theta_2 + \theta_3(2(x+1) + 1)) - (\theta_2 + \theta_3(2x + 1)) = 2\theta_3.$$
When $\theta_3$ is negative, this negative value $\theta_3$ implies the amount of change in $E[Y]$ induced by a unit change in $x$ decreases as $x$ increases: that's a constant acceleration, graphically represented by the concavity of the curve (it curls downward). When $\theta_3$ is positive, the constant acceleration corresponds to an upward-curling (or convex) curve.
For those who would like to experiment further, here is the R code that produced the data and graphics in the example.
#
# Describe the model.
#
n <- 30 # Amount of data
curvature <- -1/5 # Curvature at vertex
vertex <- 2 # Location of vertex
height <- 25 # Height of vertex
x.min <- -10 # Data extend from x=x.min to x.max
x.max <- -3
sigma <- 1 # SD of errors
#
# Simulate data.
#
set.seed(17)
epsilon <- rnorm(n, 0, sigma)
x <- seq(x.min, x.max, length.out=n)
f <- function(x) height + curvature/2 * (x - vertex)^2
#
# Plot the data, the model, and the fit.
#
X <- data.frame(x = x, y = f(x) + epsilon)
model <- lm(y ~ x + I(x^2), X)
summary(model)
library(ggplot2)
g <- ggplot(X, aes(x, y)) +
geom_smooth(method="lm", formula=y ~ x + I(x^2), size=1.5) +
geom_point()
#
# Display the plot in two ways.
#
library(grid)
dev.off()
dx <- (x.max-x.min) / (vertex+1 - x.min + x.max-x.min) # Portion for the left-hand plot
print(g + expand_limits(y=max(max(X$y), height)) +
ggtitle("Data and Fit"), vp=viewport(0, 0, dx, 1, just=c(0,0)))
print(g + stat_function(fun=f, linetype=3, size=1) +
expand_limits(x=1+vertex) +
ggtitle("Data, Fit, and True Model"),
vp=viewport(dx, 0, 1-dx, 1, just=c(0,0)))
