+23254 It appears that this is a parabola with vertex at (7.5, 4) and passing through the point (0, -3).
An equation for a parabola is y - k = a(x - h)² The vertex is (h, k)
So, we have: y - 4 = a(x - 7.5)²
To find the value for a, let's use the point (0, -3): -3 - 4 = a(0 - 7.5)²
---> -7 = a(-7.5)² ---> -7 = 56.25a ---> a = -28/225
---> y - 4 = (-28/225)(x - 7.5)²
Substituting 15 for x: y - 4 = (-28/225)(15 - 7.5)² --> y = -3
![Below is a portion of the graph of a quadratic function, $y=q(x)=ax^2+bx+c$: [asy] import graph; size(8cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.99,xmax=10.5,ymin=-5.5,ymax=5.5; pen cqcqcq=rgb(0.75,0.75,0.75); /*grid*/ pen gs=linewidth(0.7)+cqcqcq+linetype("2 2"); real gx=1,gy=1; for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs); Label laxis; laxis.p=fontsize(10); xaxis("",xmin,xmax,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); yaxis("",ymin,ymax,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); real f1(real x){return 4-(x-8)*(x-7)/8;} draw(graph(f1,-0.99,10.5),linewidth(1)); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); label("$y=q(x)$",(10.75,2.5),E); [/asy] The value of $q(15)$ is an integer. What is that integer?](/api/ssl-img-proxy?src=https%3A%2F%2Fdata.artofproblemsolving.com%2Faops20%2Flatex%2Fimages%2F86f0f95b2a2e81c932b3f2ee1abed979ccc96582.png)
