a={4,8,15,15,15,15};
f1[x_]=-2x+10;
f2[x_]=3/2x -4;
f3[x_]=-8/7 x+15*8/7;
f4[x_]=0;
f5[x_]=0;
f6[0]=0;
a[[1]]
P1=Plot[Piecewise[{{f1[x],x<=a[[1]]},{f2[x],x<=a[[2]]},{f3[x],x<=a[[3]]},{f4[x],x≤a[[4]]},{f5[x],x<=a[[5]]},{f6[x],x<=a[[6]]}}],{x,0,15}]

a={4,8,15,15,15,15};
f1[x_]=-2x+10;
f2[x_]=3/2x -4;
f3[x_]=-8/7 x+15*8/7;
f4[x_]=0;
f5[x_]=0;
f6[0]=0;
a[[1]]
P2=Plot[Piecewise[{{f1[x],x<=a[[1]]},{f2[x],x<=a[[2]]},{f3[x],x<=a[[3]]},{f4[x],x≤a[[4]]},{f5[x],x<=a[[5]]},{f6[x],x<=a[[6]]}}],{x,0,15}, PlotStyle->Red]

Show[P1,P2]

Will one polynomial of degree 5 work?

f1[x_]=x(x-4)(x-8)(x-15)

f[x_]=Integrate[f1[x],x]

Plot[-10/f[15] * f[x]+10,{x,0,15}]

-10/f[15] * f[x]+10

Degree 3 Polynomial with horizonatl tangents at end points

x0=8
y0=8
x1=15
y1=0
M={{-1/6 x0^3 + 1/2 x0^2 x1,1},{-1/6 x1^3 + 1/2 x0 x1^2,1}}
coef=Inverse[M].{y0,y1}
f[x]=coef[[1]]*(1/3 x^3 -1/2 (x0+x1) x^2 + x0 x1 x)+ coef[[2]]

Roller Coaster using cubics

a={4,8,15,15,15,15}
P2=Plot[Piecewise[{{10+(3/4) (-2 (x^2)+((x^3)/3)),x<=a[[1]]},{32-(9/16) (32 x-6 (x^2)+((x^3)/3)),x<=a[[2]]},{-(16200/343)+(48/343) (120 x-((23 (x^2))/2)+((x^3)/3)),x<=a[[3]]},{15-x,x≤a[[4]]},{x-15,x<=a[[5]]},{7,x<=a[[6]]}}],{x,0,15}]